Chapter 12: Radioactivity

In 1898, in a leaky wooden shed behind a school in Paris, Marie Curie and Pierre Curie began boiling down pitchblende — a heavy black ore left over from uranium mining. Marie had already measured something that made no sense to her. The ore was more active than the amount of uranium in it could account for. Something else was hiding inside, something she could not see, weigh out, or name.

The only way to find it was to take the ore apart, batch by batch. She stirred vats of boiling liquid with an iron rod almost as tall as she was, in a room with no proper ventilation and a stove that hardly worked. Tonnes of ore went in. Four years later, out came about a tenth of a gram of a compound of a new element: radium.

She did not need a lamp to know it was there. Left in the dark, the tubes on the shelves glowed a faint blue-green, and the Curies would come back at night simply to stand and look at them.

Nothing was burning. Nothing had been mixed in. The glow was coming from inside the atoms themselves, from a change deep in the nucleus that nobody at the time could start, stop, or slow down. This chapter is about that change, and about what comes out of a nucleus when it happens.

Diagram

Fig. 12.0-1 — Marie Curie’s Shed Laboratory Illustration of a plain wooden shed interior at night: a tall vat of ore solution on a low stove with a long iron stirring rod resting in it, sacks of black pitchblende ore stacked against the wall, and a wooden shelf holding several small sealed glass tubes that give out a faint blue-green glow. Label the sacks ‘pitchblende ore’, the vat ‘ore being separated’ and the tubes on the shelf ‘radium compound, glowing in the dark’. No flame or lamp anywhere near the tubes, so that the glow clearly comes from the tubes themselves.

NoteCuriosity Corner

Q1. Marie Curie heated, dissolved, cooled and chemically treated the ore again and again, yet its activity never weakened. Why should no chemical treatment be able to slow radioactivity down? Q2. If the radiation from a radioactive source is sent through a magnetic field, it splits into three separate paths — one bending one way, one bending the other way, and one going straight on. What could make three parts of the same beam behave so differently? Q3. Alpha particles are stopped by a sheet of paper, yet they are the most damaging of the three radiations once they get inside the body. How can the weakest penetrator be the worst offender? Q4. A nucleus contains only protons and neutrons, and no electrons at all. So where does the electron come from when a nucleus shoots one out as a beta particle? Q5. Splitting a heavy nucleus releases energy, and joining two light nuclei also releases energy. What single fact about mass could possibly explain both?

12.1 Structure Of The Atom And Nucleus

Towards the end of the nineteenth century, Sir J. J. Thomson discovered the electron, a particle far lighter than any atom and carrying a negative charge. A little later, Rutherford and others fired fast-moving alpha particles at thin metal foils and watched where they went. Most passed straight through, but a few came sharply back.

Those two pieces of work, taken together, gave us the picture of the atom used in this chapter: almost all of the mass packed into a tiny core at the centre, with the light electrons spread out around it. The picture is drawn in Fig. 12.1-1.

Diagram

Fig. 12.1-1 — General Structure Of An Atom Simple labelled sketch, not to scale: a small dense circle at the centre marked ‘nucleus (protons and neutrons)’, surrounded by three concentric circular orbits labelled K, L and M from the inside outwards, with small dots marked ‘e’ placed on the orbits. An arrow from the outermost orbit to the edge labelled ‘size of the atom, of the order of 10⁻¹⁰ m’, and an arrow to the central circle labelled ‘size of the nucleus, of the order of 10⁻¹⁵ m to 10⁻¹⁴ m’.

12.1.1 The Three Particles Of An Atom

An atom is built out of three main particles: electrons, protons and neutrons. The protons and the neutrons sit inside the nucleus, the small dense core at the centre of the atom. The electrons move around the nucleus in certain fixed orbits.

These orbits are called stationary orbits, or stationary shells. The word ‘stationary’ does not mean the electron stands still; it means that while an electron stays in one of these orbits it does not radiate out any energy. In each shell an electron has a definite energy of its own.

Charge is measured in coulomb (C). The proton carries \(+1.6 \\times 10^{-19}\\ \\text{C}\), written \(+e\). The electron carries the same size of charge with the opposite sign, \(-1.6 \\times 10^{-19}\\ \\text{C}\) or \(-e\). The neutron carries zero charge. The sign is not bookkeeping — it decides which way the particle is pushed by an electric field.

12.1.2 Shells And The Number Of Electrons They Hold

Each shell can hold only so many electrons, so an atom with more electrons needs more shells. That is why atoms of different elements have different numbers of shells.

The shells are counted outwards from the nucleus by a number \(n\), and for \(n = 1, 2, 3, 4, 5, 6, 7\) they are named K, L, M, N, O, P and Q respectively. The largest number of electrons a shell can hold is

\[\\text{maximum number of electrons in a shell} = 2n^2\]

where \(n\) is the number of that shell. Putting \(n = 1, 2, 3, \\ldots\) in turn gives 2, 8, 18, 32, 50, 72 and 98 electrons for the K, L, M, N, O, P and Q shells.

The electrons in different shells do not all have the same energy. The energy of an electron increases as the shell number \(n\) increases, so an electron in the M shell has more energy than one in the K shell.

The shells are lettered from K rather than from A. The physicist Charles Barkla, working on X-rays given out by different metals, found two clear series and named them K and L — deliberately picking letters from the middle of the alphabet, in case more series turned up later and needed names before them.

12.1.3 The Size And Contents Of The Nucleus

The size of an atom is set by the radius of the shell holding its outermost electron, and it is of the order of \(10^{-10}\\ \\text{m}\). ‘Of the order of’ is a useful phrase here: it does not claim an exact value, only that the size is around that power of ten.

The nucleus sits at the centre and is far smaller, of the order of \(10^{-15}\\ \\text{m}\) to \(10^{-14}\\ \\text{m}\), which is about \(10^{-5}\) to \(10^{-4}\) times the size of the whole atom. Almost all of the atom is empty space, which is why most of Rutherford’s alpha particles went straight through the foil.

Example 12.1 — Comparing The Atom With Its Nucleus

  1. Given: size of the atom \(\\approx 10^{-10}\\ \\text{m}\); size of the nucleus \(\\approx 10^{-15}\\ \\text{m}\). Asked: how many times bigger is the atom than its nucleus.

  2. Relation used: \(\\text{ratio} = \\dfrac{\\text{size of atom}}{\\text{size of nucleus}}\). Both lengths are already in metre, so the units cancel and the answer is a pure number with no unit.

  3. \(\\text{ratio} = \\dfrac{10^{-10}\\ \\text{m}}{10^{-15}\\ \\text{m}} = 10^{-10-(-15)} = 10^{5}\)

  4. The atom is about \(10^{5}\) times, that is one lakh times, wider than its nucleus. The answer is sensible: it is another way of saying that the nucleus is about \(10^{-5}\) times the size of the atom.

The nucleus is made of protons and neutrons. The proton carries a positive charge of \(+1.6 \\times 10^{-19}\\ \\text{C}\) and has a mass \(m\_p\) of nearly \(1.67 \\times 10^{-27}\\ \\text{kg}\). The neutron carries no charge at all, and its mass \(m\_n\) is also nearly \(1.67 \\times 10^{-27}\\ \\text{kg}\), equal to that of a proton.

The electron is the light one. Its mass \(m\_e\) is nearly \(9.1 \\times 10^{-31}\\ \\text{kg}\), which is roughly \(1/1840\) times the mass of a proton. These figures are collected in Table 12.1.

Table 12.1 — Symbol, charge and mass of the electron, proton and neutron

Parameter Electron Proton Neutron
Symbol \(e\) \(p\) \(n\)
Charge \(-1.6 \\times 10^{-19}\\ \\text{C}\) \(+1.6 \\times 10^{-19}\\ \\text{C}\) zero
Mass \(9.1 \\times 10^{-31}\\ \\text{kg}\) \(1.67 \\times 10^{-27}\\ \\text{kg}\) \(1.67 \\times 10^{-27}\\ \\text{kg}\)

Measured more closely, \(m\_p = 1.6726 \\times 10^{-27}\\ \\text{kg}\) and \(m\_n = 1.6749 \\times 10^{-27}\\ \\text{kg}\), so the neutron is very slightly the heavier of the two.

Nucleons And The Mass Of The Atom

Protons and neutrons are the main constituents of the nucleus, and together they are called nucleons. If a nucleus holds \(Z\) protons, its total charge is \(+Ze\), since the neutrons add no charge. The mass of the nucleus is roughly the mass of one proton or one neutron multiplied by the total number of nucleons.

Because an electron is thousands of times lighter than a nucleon, the electrons add almost nothing to the mass of an atom. The total mass of an atom can therefore be taken to be the same as the mass of its nucleus.

12.1.4 Atomic Number And Mass Number

Two counts fix which atom we are dealing with. The first is the number of protons in the nucleus. This number decides the place of the atom in the periodic table, and it is called the atomic number, denoted by \(Z\).

NoteDefinition

Atomic Number (Z) The atomic number of an atom is the number of protons in its nucleus, which is the same as the number of electrons in a neutral atom. It is a pure count and has no unit. \[Z = \\text{number of protons in the nucleus of an atom}\]

Q. A neutral atom of an element has 8 electrons outside its nucleus. What is its atomic number?

A. 8. In a neutral atom the number of electrons equals the number of protons, and the atomic number counts the protons.

The second count is the total number of nucleons — protons and neutrons together. This is called the mass number, denoted by \(A\).

NoteDefinition

Mass Number (A) The mass number of an atom is the total number of nucleons in its nucleus, that is, the sum of the number of protons and the number of neutrons. It is a pure count and has no unit. \[A = \\text{number of protons} + \\text{number of neutrons in the nucleus}\]

Since \(A\) counts protons and neutrons together while \(Z\) counts only the protons, subtracting one from the other leaves the neutrons:

\[\\text{number of neutrons} = A - Z\]

Atomic number \(Z\) counts protons only. Mass number \(A\) counts protons and neutrons together. The one distinction: \(Z\) tells you which element it is, \(A\) tells you how heavy that particular nucleus is. Neither carries a unit.

12.2 The Atomic Model

12.2.1 Counting The Particles In A Neutral Atom

An atom as a whole shows no charge. Since each proton carries \(+e\) and each electron carries \(-e\), the only way for the charges to cancel is for the two to be equal in number. So in any neutral atom, the number of electrons revolving around the nucleus equals the number of protons inside it.

That single fact, together with the meaning of \(A\) and \(Z\), fixes the whole count. If an atom has atomic number \(Z\) and mass number \(A\), then

\[\\text{number of electrons} = Z,\\qquad \\text{number of protons} = Z,\\qquad \\text{number of neutrons} = A - Z\]

An atom is specified in writing by the symbol

\[^{A}\_{Z}\\mathrm{X}\]

where X is the chemical symbol of the element, \(A\) is written above and to the left as the mass number, and \(Z\) below and to the left as the atomic number. Reading such a symbol is simply reading those two counts off the page.

The SI unit of energy is the joule (J). Inside atoms the joule is far too large, so energy is quoted in electron volt (eV) and in mega electron volt (MeV), where \(1\\ \\text{MeV} = 10^{6}\\ \\text{eV}\). Both are units of energy, not of charge or voltage.

12.2.2 Models Of Hydrogen, Helium And Sodium

An atom is far too small to see — about \(10^{-10}\\ \\text{m}\) across. To picture how its parts are arranged, we draw a model of it, and such a model is never to scale: the nucleus is drawn far larger than it really is compared with the shells.

The lightest atom is hydrogen, with \(A = 1\) and \(Z = 1\). It is written \(^{1}\_{1}\\mathrm{H}\). It has one proton in its nucleus and one electron in the K shell, as shown in Fig. 12.2-1. Its number of neutrons is \(A - Z = 1 - 1 = 0\).

Diagram

Fig. 12.2-1 — Model Of The Hydrogen Atom Not to scale. A small central circle labelled ‘1p’, with one circular orbit around it labelled K carrying a single dot marked ‘e’. Below the sketch write the symbol ¹₁H and the word ‘Hydrogen’.

The helium atom has \(A = 4\) and \(Z = 2\), written \(^{4}\_{2}\\mathrm{He}\). It has 2 protons inside the nucleus and 2 electrons outside in the K shell, and its number of neutrons is \(A - Z = 4 - 2 = 2\). Its model is drawn in Fig. 12.2-2.

Diagram

Fig. 12.2-2 — Model Of The Helium Atom Not to scale. A central circle labelled ‘2p, 2n’, with one circular orbit around it labelled K carrying two dots marked ‘e’. Below the sketch write the symbol ⁴₂He and the word ‘Helium’.

The sodium atom has \(Z = 11\) and \(A = 23\), written \(^{23}\_{11}\\mathrm{Na}\). Its nucleus holds 11 protons and \(A - Z = 23 - 11 = 12\) neutrons, and its 11 electrons are shared out over three shells — 2 in the K shell, 8 in the L shell and 1 in the M shell, as drawn in Fig. 12.2-3.

Diagram

Fig. 12.2-3 — Model Of The Sodium Atom Not to scale. A central circle labelled ‘11p, 12n’, with three concentric orbits around it labelled K, L and M from inside outwards. Place 2 dots marked ‘e’ on the K orbit, 8 dots marked ‘e’ on the L orbit and 1 dot marked ‘e’ on the M orbit. Below the sketch write the symbol ²³₁₁Na and the word ‘Sodium’.

An atom can now be photographed. A single strontium atom, held still in an electric field, has been captured in a photograph. Its nucleus contains 38 protons, and the atom itself is only a few millionths of a millimetre across.

Example 12.2 — Reading A Nuclide Symbol

  1. Given: an aluminium atom written as \(^{27}\_{13}\\mathrm{Al}\), so mass number \(A = 27\) and atomic number \(Z = 13\). Asked: the number of protons, neutrons and electrons, and how the electrons are shared out among the shells.

  2. Relations used: number of protons \(= Z\); number of electrons \(= Z\) for a neutral atom; number of neutrons \(= A - Z\). The shells are filled outwards from K, and the K, L and M shells hold at most 2, 8 and 18 electrons.

  3. Number of protons \(= 13\). Number of electrons \(= 13\). Number of neutrons \(= A - Z = 27 - 13 = 14\). Filling outwards: 2 electrons in K, 8 in L, and the remaining \(13 - 2 - 8 = 3\) in M.

  4. The atom has 13 protons and 14 neutrons in its nucleus, and 13 electrons arranged as 2, 8, 3 — that is, 2 in K, 8 in L and 3 in M. These are counts, so they carry no unit. The check: protons and neutrons add to \(13 + 14 = 27\), which is the mass number, as it must be.

12.2.3 Chemical Change Compared With Nuclear Change

There are two quite different ways an atom can change, and telling them apart matters for everything that follows in this chapter.

In a chemical change, only the orbital electrons are affected. Atoms share or transfer their outer electrons, and the nucleus is left untouched. In a nuclear change, the number of nucleons inside the nucleus itself changes.

The two also sit at completely different energy scales. A chemical change involves energies of the order of a few eV, while a nuclear change involves energies of the order of a few MeV — nearly \(10^{6}\) times as much. The comparison is set out in Table 12.2.

Table 12.2 — Chemical Change Compared With Nuclear Change

Property Chemical change Nuclear change
What changes The number of orbital electrons of the atom The number of nucleons inside the nucleus
Part of the atom involved The outer shells The nucleus
Energy involved Low, of the order of a few eV High, of the order of a few MeV
Does the element change? No, the nucleus is untouched Yes, a nucleus of a new element may be formed
What Is Conserved In A Nuclear Reaction

In a nuclear reaction, two counts are always conserved. The total atomic number — the total number of protons — is the same before and after, and so is the total mass number, the total number of protons and neutrons.

In other words, the sum of the atomic numbers of the reactants equals the sum of the atomic numbers of the products, and the sum of their mass numbers likewise. Every nuclear equation later in this chapter can be checked against these two rules, and none of them may break either one.

A neutral atom is written as \(^{40}\_{18}\\mathrm{Ar}\). State the number of protons, the number of neutrons and the number of electrons it has, and write down how its electrons are distributed among the K, L and M shells.

12.3 Isotopes

12.3.1 What Isotopes Are

Not every atom of an element has the same mass. Two atoms of the same element can have nuclei of different masses, because they hold different numbers of neutrons while holding the same number of protons.

NoteDefinition

Isotopes The atoms of the same element, having the same atomic number \(Z\) but different mass number \(A\), are called isotopes. Both \(Z\) and \(A\) are counts, so neither carries a unit.

Since the place of an element in the periodic table is decided by \(Z\) alone, all the isotopes of an element occupy the same place in the table. That is where the name comes from.

The atoms of isotopes have the same number of protons \(Z\), but a different number of neutrons \(A - Z\), in their nuclei. Being neutral atoms, they also have the same number of electrons outside the nucleus — and since chemical behaviour is decided by those outer electrons, the chemical properties of all the isotopes of an element are the same.

The three isotopes of hydrogen are named from the Greek words for first, second and third — protos, deuteros and tritos — which is how protium, deuterium and tritium got their names. The naming records nothing about their chemistry, only how many nucleons each nucleus carries.

In nature, the different isotopes of an element occur in different proportions. In most cases one isotope is far more abundant than the rest, and the element is commonly specified by the mass number of that most abundant isotope.

12.3.2 Isotopes Of Hydrogen, Carbon And Chlorine

Hydrogen has three isotopes: protium \(^{1}*{1}\\mathrm{H}\), which is ordinary hydrogen and by far the most abundant; deuterium \(^{2}*{1}\\mathrm{H}\), also called heavy hydrogen; and tritium \(^{3}\_{1}\\mathrm{H}\). Each has one proton in its nucleus, so \(Z = 1\) for all three, and each has one electron outside the nucleus.

Their nuclei differ only in neutrons. Protium has none, deuterium has one and tritium has two, as shown in Fig. 12.3-1 and set out in Table 12.3.

Diagram

Fig. 12.3-1 — Nuclei Of The Three Isotopes Of Hydrogen Three small labelled sketches side by side, not to scale. (a) a nucleus containing one proton only, marked ‘1p’, labelled ¹₁H Protium. (b) a nucleus containing one proton and one neutron, marked ‘1p, 1n’, labelled ²₁H Deuterium. (c) a nucleus containing one proton and two neutrons, marked ‘1p, 2n’, labelled ³₁H Tritium. Show one electron on a single K orbit around each nucleus.

Table 12.3 — Protons and neutrons in the isotopes of hydrogen

Parameter Protium \(^{1}\_{1}\\mathrm{H}\) Deuterium \(^{2}\_{1}\\mathrm{H}\) Tritium \(^{3}\_{1}\\mathrm{H}\)
Values of \(Z\) and \(A\) \(Z = 1,\\ A = 1\) \(Z = 1,\\ A = 2\) \(Z = 1,\\ A = 3\)
Number of protons 1 1 1
Number of neutrons 0 1 2

Carbon has three isotopes, \(^{12}*{6}\\mathrm{C}\), \(^{13}*{6}\\mathrm{C}\) and \(^{14}*{6}\\mathrm{C}\), of which \(^{12}*{6}\\mathrm{C}\) is the most abundant. Each has 6 protons in its nucleus, so \(Z = 6\) throughout, and 6 electrons outside it. Their neutron counts are 6, 7 and 8 respectively.

Chlorine has two isotopes, \(^{35}*{17}\\mathrm{Cl}\) and \(^{37}*{17}\\mathrm{Cl}\), whose relative abundance in nature is in the ratio \(3 : 1\). Each has 17 protons in its nucleus and 17 electrons outside it, while \(^{35}*{17}\\mathrm{Cl}\) has 18 neutrons and \(^{37}*{17}\\mathrm{Cl}\) has 20 neutrons.

Example 12.3 — Testing Whether Two Nuclei Are Isotopes

  1. Given: two nuclei, \(^{16}*{8}\\mathrm{O}\) and \(^{18}*{8}\\mathrm{O}\). Asked: are they isotopes, and how many neutrons does each contain?

  2. Relations used: two nuclei are isotopes if their atomic numbers \(Z\) are equal while their mass numbers \(A\) differ; and the number of neutrons is \(A - Z\).

  3. For both nuclei \(Z = 8\), while \(A = 16\) for the first and \(A = 18\) for the second. Same \(Z\), different \(A\), so they are isotopes. Neutrons in the first \(= 16 - 8 = 8\). Neutrons in the second \(= 18 - 8 = 10\).

  4. They are isotopes of oxygen, with 8 and 10 neutrons respectively. These are counts, so they carry no unit. The check: the two nuclei differ by exactly 2 neutrons, and their mass numbers differ by 2, as they must.

Look up chlorine in a periodic table. Its atomic mass is printed as about 35.5, which is not a whole number, although every individual nucleus has a whole number of nucleons. Using the fact that \(^{35}*{17}\\mathrm{Cl}\) and \(^{37}*{17}\\mathrm{Cl}\) occur in the ratio \(3 : 1\), work out why 35.5 is the figure that appears.

12.3.3 Stable And Radioactive Isotopes

Isotopes are of two kinds. Stable isotopes are those in which the number of neutrons is nearly equal to the number of protons in the nucleus. Unstable, or radioactive isotopes, are those in which the number of neutrons is much greater than the number of protons. These undergo radioactive decay, and they are of great medical and industrial use.

Take uranium. In \(^{235}*{92}\\mathrm{U}\) and \(^{238}*{92}\\mathrm{U}\) each nucleus has 92 protons, but \(^{235}*{92}\\mathrm{U}\) has \(235 - 92 = 143\) neutrons while \(^{238}*{92}\\mathrm{U}\) has \(238 - 92 = 146\) neutrons. In both, the neutrons far outnumber the protons, so both are radio isotopes.

The same test picks out \(^{14}\_{6}\\mathrm{C}\) from among the three isotopes of carbon. With 6 protons and 8 neutrons it is the one carrying the largest excess of neutrons, and it is the radio isotope of the three.

Among all the elements, tin has ten stable isotopes, more than any other element.

The two nuclei \(^{58}*{28}\\mathrm{Ni}\) and \(^{60}*{28}\\mathrm{Ni}\) occur in the same element.

(i) State, with a reason, whether they are isotopes.

(ii) Find the number of neutrons in each nucleus.

(iii) State how many electrons a neutral atom of each has.

12.4 Isobars

Isotopes were atoms of one element with different masses. Now take the opposite case: two atoms of different elements that happen to carry the same total number of nucleons.

NoteDefinition

Isobars The atoms of different elements which have the same mass number \(A\), but different atomic number \(Z\), are called isobars. Both are counts and carry no unit.

The atoms of isobars have the same number of nucleons \(A\) in their nuclei, but a different number of protons \(Z\) and hence a different number of neutrons \(A - Z\). And since the number of electrons outside the nucleus is always equal to the number of protons, isobars also have different numbers of electrons.

The name comes from the Greek words isos, meaning equal, and baros, meaning weight — atoms of equal weight. The same word appears on a weather map, where an isobar is a line joining places of equal pressure. The Greek root is doing the same job in both subjects.

Two Pairs Of Isobars

Sodium \(^{23}*{11}\\mathrm{Na}\) and magnesium \(^{23}*{12}\\mathrm{Mg}\) are isobars. The sodium nucleus contains 11 protons and 12 neutrons, with 11 electrons outside it; the magnesium nucleus contains 12 protons and 11 neutrons, with 12 electrons outside it. The total number of protons and neutrons is 23 in each, which is why they are isobars. The two nuclei are drawn side by side in Fig. 12.4-1.

Diagram

Fig. 12.4-1 — Nuclei Of A Pair Of Isobars Two labelled circles side by side, not to scale. The left circle is marked ‘11p, 12n’ and labelled ²³₁₁Na; the right circle is marked ‘12p, 11n’ and labelled ²³₁₂Mg. Under each circle write ‘total nucleons = 23’. Between the two circles place a caption arrow labelled ‘same A, different Z’.

Carbon \(^{14}*{6}\\mathrm{C}\) and nitrogen \(^{14}*{7}\\mathrm{N}\) are also isobars. The carbon nucleus has 6 protons and 8 neutrons, with 6 electrons outside it, while the nitrogen nucleus has 7 protons and 7 neutrons, with 7 electrons outside it. Again the nucleon totals match at 14, while everything else differs.

When the numbers of protons and neutrons in one nucleus are exactly interchanged in the other, as in \(^{23}*{11}\\mathrm{Na}\) and \(^{23}*{12}\\mathrm{Mg}\), the pair is called a pair of mirror isobars.

Isotopes have the same \(Z\) and different \(A\); they are the same element. Isobars have the same \(A\) and different \(Z\); they are different elements. The single distinction: isotopes share the number of protons, isobars share only the total number of nucleons.

Example 12.4 — Testing Whether Two Nuclei Are Isobars

  1. Given: two nuclei, \(^{50}*{22}\\mathrm{Ti}\) and \(^{50}*{24}\\mathrm{Cr}\). Asked: are they isobars, and how many neutrons does each contain?

  2. Relations used: two nuclei are isobars if their mass numbers \(A\) are equal while their atomic numbers \(Z\) differ; and the number of neutrons is \(A - Z\).

  3. For both nuclei \(A = 50\), while \(Z = 22\) for the first and \(Z = 24\) for the second. Same \(A\), different \(Z\), so they are isobars. Neutrons in the first \(= 50 - 22 = 28\). Neutrons in the second \(= 50 - 24 = 26\).

  4. They are isobars, with 28 and 26 neutrons respectively — counts, so no unit. The check: the second nucleus has 2 protons more and 2 neutrons fewer, so its nucleon total is unchanged at 50, exactly as an isobar requires.

Consider the three nuclei \(^{40}*{19}\\mathrm{K}\), \(^{41}*{19}\\mathrm{K}\) and \(^{40}\_{20}\\mathrm{Ca}\).

(i) Name the pair which are isotopes, and give a reason.

(ii) Name the pair which are isobars, and give a reason.

(iii) State the number of neutrons in each of the three nuclei.

12.5 Isotones

Isotopes matched in protons and isobars matched in total nucleons. There is one more way two nuclei can match: they can hold the same number of neutrons.

NoteDefinition

Isotones The atoms having a different number of protons but the same number of neutrons — that is, different \(Z\) and different \(A\), but the same value of \(A - Z\) — are called isotones. They have different numbers of electrons.

Since the number of protons decides the element, isotones are always atoms of different elements. And because the number of electrons in a neutral atom equals the number of protons, isotones differ in their electrons too, and so in their chemistry.

Q. Two nuclei have the same number of neutrons. Must they have the same mass number?

A. No. Their proton counts differ, and \(A\) is protons plus neutrons, so their mass numbers differ as well.

Two Pairs Of Isotones

Sodium \(^{23}*{11}\\mathrm{Na}\) and magnesium \(^{24}*{12}\\mathrm{Mg}\) are isotones. The sodium nucleus has \(23 - 11 = 12\) neutrons and the magnesium nucleus has \(24 - 12 = 12\) neutrons, so the neutron counts match at 12, while sodium has 11 protons and magnesium has 12. The two nuclei are compared in Fig. 12.5-1.

Diagram

Fig. 12.5-1 — Nuclei Of A Pair Of Isotones Two labelled circles side by side, not to scale. The left circle is marked ‘11p, 12n’ and labelled ²³₁₁Na; the right circle is marked ‘12p, 12n’ and labelled ²⁴₁₂Mg. Shade or tint the neutrons in both circles in the same colour and add the caption arrow between them labelled ‘same number of neutrons, different Z and A’.

Potassium \(^{39}*{19}\\mathrm{K}\) and calcium \(^{40}*{20}\\mathrm{Ca}\) are also isotones, each nucleus having 20 neutrons.

A quick way to keep the three names apart is to read the letter inside each: isotopes have the same number of protons, isobars have the same mass number A, and isotones have the same number of neutrons.

12.5.1 Isotopes, Isobars And Isotones Compared

The three kinds of relationship are set apart by which of the three counts — \(Z\), \(A\) or \(A - Z\) — is the same, as summarised in Table 12.4.

Table 12.4 — Isotopes, isobars and isotones compared

Parameter Isotope Isobar Isotone
\(Z\) (number of protons) same different different
\(A\) (number of nucleons) different same different
\(A - Z\) (number of neutrons) different different same

Reading a row across is a useful check on your own work: in any correctly identified pair, exactly one of these three counts matches and the other two do not.

Example 12.5 — Sorting Three Nuclei

  1. Given: the three nuclei \(^{19}*{9}\\mathrm{F}\), \(^{20}*{10}\\mathrm{Ne}\) and \(^{22}*{10}\\mathrm{Ne}\). Asked: which pairs are isotopes and which are isotones.

  2. Relations used: isotopes have the same \(Z\); isotones have the same \(A - Z\). So the neutron count \(A - Z\) must be worked out for each nucleus before the pairs can be sorted.

  3. Neutrons in \(^{19}*{9}\\mathrm{F} = 19 - 9 = 10\). Neutrons in \(^{20}*{10}\\mathrm{Ne} = 20 - 10 = 10\). Neutrons in \(^{22}*{10}\\mathrm{Ne} = 22 - 10 = 12\). Comparing \(Z\): the two neon nuclei share \(Z = 10\), so they are isotopes. Comparing \(A - Z\): fluorine and \(^{20}*{10}\\mathrm{Ne}\) both have 10 neutrons, so they are isotones.

  4. \(^{20}*{10}\\mathrm{Ne}\) and \(^{22}*{10}\\mathrm{Ne}\) are isotopes; \(^{19}*{9}\\mathrm{F}\) and \(^{20}\_{10}\\mathrm{Ne}\) are isotones. All the figures are counts and carry no unit. The check: no two of the three nuclei share a mass number, so there is no isobar pair here, which agrees with the answer.

You are given four nuclei: \(^{12}*{6}\\mathrm{C}\), \(^{13}*{6}\\mathrm{C}\), \(^{14}*{7}\\mathrm{N}\) and \(^{13}*{7}\\mathrm{N}\). Pick out one pair of isotopes, one pair of isobars and one pair of isotones, giving in each case the count that the pair has in common.

12.6 Radioactivity

12.6.1 Becquerel And The Fogged Photographic Plate

The phenomenon of radioactivity was discovered by Henry Becquerel in 1896. In a dark room he left a uranium salt, wrapped in black paper, lying on a photographic plate. A photographic plate kept in the dark and never exposed to light should stay blank when it is developed.

After some days he developed the plate and found it had been affected all the same. He repeated the experiment with other salts of uranium and got the same result each time. The arrangement is shown in Fig. 12.6-1.

Diagram

Fig. 12.6-1 — Becquerel’s Observation Two panels side by side. Left panel: a photographic plate lying flat in a dark closed drawer, with a small heap of uranium salt resting on it, the salt fully wrapped in a sheet of black paper; label the plate ‘photographic plate’, the wrapping ‘black paper cover’ and the heap ‘uranium salt’; draw short arrows from the salt down through the paper to the plate labelled ‘radiations’. Right panel: the same plate after developing, showing a dark patch exactly under where the salt had lain, labelled ‘plate affected’.

From these observations Becquerel concluded that uranium and its salts, by themselves, give out some kind of radiations. These radiations can pass through the cover — black paper, glass or wood — and still affect the plate underneath. They were called the Becquerel rays.

The SI unit in which the activity of a radioactive source is measured is named the becquerel (Bq) after him. One becquerel means one nucleus disintegrating per second, so it is a rate of decay rather than an amount of substance.

12.6.2 Radioactive Substances

On further investigation, the Becquerel rays turned out not to be one thing but three. Some of the radiation was positively charged, some negatively charged, and some carried no charge at all. These three were named alpha (\(\\alpha\)), beta (\(\\beta\)) and gamma (\(\\gamma\)) radiations respectively. The substances giving them out were called radioactive substances.

NoteDefinition

Radioactive Substances The substances which disintegrate, or decay, by the spontaneous emission of radiations are called radioactive substances — for example uranium, radium, polonium, thorium and actinium.

The word ‘spontaneous’ is doing real work in that definition. Nobody starts the process off and nobody keeps it going; the nucleus emits when it emits.

The isotopes of nearly all the elements of atomic number higher than 82 — that is, everything after lead in the periodic table — are radioactive, because in their nuclei the number of neutrons is much greater than the number of protons. These are called the natural radioactive substances.

Example 12.6 — Picking Out The Natural Radioactive Substances

  1. Given: three nuclei, \(^{207}*{82}\\mathrm{Pb}\), \(^{226}*{88}\\mathrm{Ra}\) and \(^{232}*{90}\\mathrm{Th}\). Asked: which of them are natural radioactive substances.

  2. Rule used: the isotopes of nearly all elements with atomic number higher than 82 are radioactive. So the test is on \(Z\) alone, and \(Z\) must be read off each symbol.

  3. For \(^{207}*{82}\\mathrm{Pb}\), \(Z = 82\), which is not higher than 82. For \(^{226}*{88}\\mathrm{Ra}\), \(Z = 88\), which is higher than 82. For \(^{232}*{90}\\mathrm{Th}\), \(Z = 90\), which is higher than 82.

  4. Radium and thorium are natural radioactive substances; lead is not. A check on the answer: lead is the element at which the rule stops, and it is exactly the element that the heavy radioactive nuclei end up decaying into.

12.6.3 Radioactivity Is A Nuclear Phenomenon

Where in the atom does this radiation come from — the electrons outside, or the nucleus within? The answer was settled by observation, not by argument.

A radioactive substance was subjected to every kind of physical change, such as a change in pressure or temperature, and every kind of chemical change, such as excessive heating, freezing, the action of strong electric and magnetic fields, chemical treatment and oxidation. None of these changed the nature of the radiation emitted, and none changed its rate of decay.

Now, all of those treatments act on the orbital electrons of an atom — that is precisely how chemistry works. If radioactivity were a property of the orbital electrons, it would have been easily affected by such changes. Since it was not affected at all, radioactivity must be a property of the nucleus.

NoteDefinition

Radioactivity Radioactivity is a nuclear phenomenon. It is the process of spontaneous emission of \(\\alpha\) or \(\\beta\) and \(\\gamma\) radiations from the nucleus of atoms during their decay.

Q1. Marie Curie heated, dissolved, cooled and chemically treated the ore again and again, yet its activity never weakened. Why should no chemical treatment be able to slow radioactivity down?

A. Because chemical treatment only rearranges the outer electrons, while radioactivity comes from the nucleus. Nothing she could do in a vat reached that deep, so the activity stayed exactly as it was.

Decay Is A Random Process

A radioactive substance contains an enormous number of atoms. There is no way to know, or to predict, when any one nucleus will decay, or which nucleus will be the next to do so.

Radioactive decay is therefore a random phenomenon. There is no law by which the disintegration of an individual nucleus can be known or predicted in advance.

A student claims that a radioactive salt will decay faster if it is heated strongly in a flame, and slower if it is cooled in ice.

(i) State whether the claim is correct.

(ii) Give a reason for your answer, naming the part of the atom involved.

12.7 Separating The Alpha, Beta And Gamma Radiations

Naming three radiations is one thing; showing that a real source gives out three is another. In 1903 Rutherford studied the nature of the radiations emitted by radioactive substances experimentally, and the way he separated them out is still the clearest demonstration available.

In both the experiments that follow, the radioactive substance is placed at the bottom of a thick lead box with a narrow opening at the top. Lead absorbs the radiations striking its walls from inside, so only a narrow beam escapes through the opening — and a narrow beam is what is needed if separate paths are to be told apart.

12.7.1 Deflection In A Magnetic Field

The beam from the lead box is passed through a magnetic field acting perpendicular to its path. Here the field is directed normal to the plane of the paper, inwards, and the paths that result are drawn in Fig. 12.7-1.

Diagram

Fig. 12.7-1 — Deflection Of Radioactive Radiations In A Magnetic Field A thick-walled lead box, hatched, open at the top, with a small dot at its base labelled ‘radioactive substance’; label the walls ‘lead box’. Above the opening, fill the region with a grid of small crosses labelled ‘magnetic field inwards, normal to the plane of the paper’. From the opening draw three paths: one curving to the left labelled α, one curving more sharply to the right labelled β, and one going straight up the middle labelled γ.

The three paths tell us the charge on each radiation. Those which turn to the left must be positively charged, and these are the alpha particles. Those which turn to the right must be negatively charged, and these are the beta particles. The direction of deflection in a magnetic field follows from Fleming’s left hand rule.

The \(\\beta\) particles are deviated more than the \(\\alpha\) particles. A magnetic force of a given size bends a light particle through a larger angle than a heavy one, and a \(\\beta\) particle is far lighter than an \(\\alpha\) particle.

The third path shows no deflection at all. Radiations that pass straight on must be uncharged, and these are the gamma radiations. Gamma radiations are electromagnetic waves, similar to light waves, and being uncharged they are not affected by the magnetic field.

Fleming’s left hand rule was met earlier, for the force on a current-carrying conductor in a magnetic field. A stream of charged particles is a current, so the same rule applies here — with the reminder that conventional current is taken along the motion of positive charge, and opposite to the motion of the \(\\beta\) particles.

12.7.2 Deflection In An Electric Field

The same beam can be sent instead through an electric field acting perpendicular to its path, set up between two oppositely charged plates. Once again it separates into three constituents, as shown in Fig. 12.7-2.

Diagram

Fig. 12.7-2 — Deflection Of Radioactive Radiations In An Electric Field A thick-walled lead box, hatched, open at the top, with a dot at its base labelled ‘radioactive substance’. Above it, two vertical plates face each other, the left plate marked ‘−’ and the right plate marked ‘+’, with a horizontal arrow across the gap labelled ‘electric field’. From the opening draw three paths: one bending towards the left-hand negative plate labelled α, one bending more sharply towards the right-hand positive plate labelled β, and one going straight up between them labelled γ.

Those which turn towards the negative plate are the positively charged alpha particles, since unlike charges attract. Those which turn towards the positive plate are the negatively charged beta particles. Those which pass undeviated are the uncharged gamma radiations. As in the magnetic field, the beta particles are deviated more than the alpha particles.

Which way a radiation bends and how far it bends are two separate facts. The sign of the charge decides the direction — \(\\alpha\) one way, \(\\beta\) the opposite way, \(\\gamma\) not at all. The mass decides the amount, and the lighter \(\\beta\) particle is always deflected the more of the two.

Q2. If the radiation from a radioactive source is sent through a magnetic field, it splits into three separate paths — one bending one way, one bending the other way, and one going straight on. What could make three parts of the same beam behave so differently?

A. Their charges. The beam is really three different radiations mixed together: positively charged \(\\alpha\) particles, negatively charged \(\\beta\) particles, which are pushed the opposite way, and uncharged \(\\gamma\) radiations, which feel no force at all and carry straight on.

Example 12.7 — Naming The Radiations From Their Paths

  1. Given: a beam from a radioactive source passes between two plates, one positive and one negative, and splits into three paths — path A bends towards the positive plate, path B goes straight through, and path C bends towards the negative plate. Asked: name the radiation along each path, and state which of A and C is deflected more.

  2. Reasoning used: a charge is attracted towards the plate of opposite sign, and an uncharged radiation is undeflected. For the amount of deflection, the lighter particle bends more.

  3. Path A is attracted towards the positive plate, so it carries negative charge — these are \(\\beta\) particles. Path C is attracted towards the negative plate, so it carries positive charge — these are \(\\alpha\) particles. Path B is undeflected, so it is uncharged — these are \(\\gamma\) radiations. Of A and C, the \(\\beta\) particles of path A are far lighter, so A is deflected more.

  4. A is \(\\beta\), B is \(\\gamma\), C is \(\\alpha\), and A shows the larger deflection. The check: the two deflected paths bend towards opposite plates, which is what oppositely charged particles must do.

A radioactive source is placed inside a thick lead box with a narrow opening, and the beam leaving the opening is passed through a magnetic field directed into the plane of the paper.

(i) State why the source is kept inside a lead box with only a narrow opening.

(ii) Name the radiation that passes through the field undeviated, and give a reason.

(iii) Name the radiation that is deflected the most, and give a reason.

12.8 Properties Of Alpha Particles

What An Alpha Particle Is

An alpha particle consists of two protons and two neutrons. That is exactly the make-up of a helium nucleus, so an \(\\alpha\) particle is the same thing as a doubly ionised helium atom — a helium atom that has lost both its orbital electrons. It is represented as \(^{4}\_{2}\\mathrm{He}\) or as \(\\mathrm{He}^{2+}\).

The difference between a helium atom and an \(\\alpha\) particle is only the two electrons, as Fig. 12.8-1 shows.

Diagram

Fig. 12.8-1 — Helium Atom And Alpha Particle Two sketches side by side, not to scale. (a) A central circle marked ‘2p, 2n’ with one circular orbit around it carrying two dots marked ‘e’; label it ‘helium atom’. (b) The same central circle marked ‘2p, 2n’ with no orbit and no electrons around it; label it ‘alpha particle’.

If an \(\\alpha\) particle (\(\\mathrm{He}^{2+}\)) gains one electron it becomes a singly ionised helium atom (\(\\mathrm{He}^{+}\)), and if it gains two electrons it becomes a neutral helium atom (He).

Mass, Charge And Specific Charge

The mass of an \(\\alpha\) particle is roughly four times the mass of a proton, that is \(4m\_p\), which comes to \(6.68 \\times 10^{-27}\\ \\text{kg}\). Its charge is twice the charge of a proton, that is \(+3.2 \\times 10^{-19}\\ \\text{C}\), or \(+2e\).

Charge and mass are often combined into one quantity, the specific charge, which is simply the charge carried per kilogram of the particle:

\[\\text{specific charge} = \\frac{q}{m}\]

where \(q\) is the charge in coulomb (C), \(m\) is the mass in kilogram (kg), and the specific charge is therefore in coulomb per kilogram (\(\\text{C kg}^{-1}\)). For the \(\\alpha\) particle its value is \(4.79 \\times 10^{7}\\ \\text{C kg}^{-1}\).

specific charge = q ÷ m It says how much charge each kilogram of the particle carries. A big value means a light particle carrying a lot of charge, which is exactly the kind that a field pushes about most easily. Its unit is C kg⁻¹.

Example 12.8 — Specific Charge Of An Alpha Particle

  1. Given: charge on an \(\\alpha\) particle \(q = 3.2 \\times 10^{-19}\\ \\text{C}\); mass of an \(\\alpha\) particle \(m = 6.68 \\times 10^{-27}\\ \\text{kg}\). Asked: its specific charge.

  2. Relation used: \(\\text{specific charge} = \\dfrac{q}{m}\), since specific charge is defined as charge per unit mass. Both quantities are already in SI units, so no conversion is needed.

  3. \(\\dfrac{q}{m} = \\dfrac{3.2 \\times 10^{-19}\\ \\text{C}}{6.68 \\times 10^{-27}\\ \\text{kg}} = 0.479 \\times 10^{8}\\ \\text{C kg}^{-1}\)

  4. The specific charge of an \(\\alpha\) particle is \(4.79 \\times 10^{7}\\ \\text{C kg}^{-1}\). The unit works out as coulomb per kilogram, as it must for a charge divided by a mass.

Speed

The speed of \(\\alpha\) particles is of the order of \(10^{7}\\ \\text{m s}^{-1}\). It is not one fixed value: \(\\alpha\) particles emitted from different radioactive substances travel at different speeds, and even those emitted from the nuclei of the same substance do not all carry the same energy, though their energies lie within a small range.

Ionising Power

An \(\\alpha\) particle strongly ionises the gas through which it passes.

NoteDefinition

Ionisation When an energetic particle collides with an atom of a gas, it can knock an electron out of that atom, and the atom is left as a positive ion. This process is called ionisation.

The ionising power of \(\\alpha\) particles is roughly 100 times that of \(\\beta\) particles and roughly 10,000 times that of \(\\gamma\) radiations. It is the largest of the three.

Ionising power is what makes a household smoke alarm work. Inside one type of alarm, a small sealed source emits \(\\alpha\) particles that ionise the air in a tiny chamber, and the ions allow a small steady current to flow across it. Smoke particles drifting into the chamber soak up the ions, the current drops, and the drop is what sets off the alarm. The source is chosen deliberately: \(\\alpha\) particles are stopped by the plastic casing, so nothing harmful gets out, which is also why such an alarm must never be opened up or taken apart.

Penetrating Power

Every ionisation costs the \(\\alpha\) particle some of its energy, and since it ionises so heavily it rapidly dissipates its energy as it moves through a medium. Its penetrating power is therefore quite small.

An \(\\alpha\) particle can penetrate only through 3 to 8 cm in air, and it can easily be stopped by a thin card sheet or a thick paper. Its penetrating power is roughly \(1/100\) times that of a \(\\beta\) particle and \(1/10{,}000\) times that of \(\\gamma\) radiation.

Ionising power is how much damage a radiation does along its path; penetrating power is how far it gets before it stops. They run opposite to each other. The \(\\alpha\) particle has the maximum ionising power and the least penetrating power — because heavy ionising is what uses its energy up.

Behaviour In Electric And Magnetic Fields

Alpha particles are positively charged, so they are deflected by electric and magnetic fields. The deflection of an \(\\alpha\) particle is less than that of a \(\\beta\) particle, because the \(\\alpha\) particle has the greater mass.

Other Properties

Alpha particles affect a photographic plate — the very effect by which radioactivity was first noticed.

They cause fluorescence on striking a fluorescent material, that is, the material gives out a small flash of light where the particle lands.

They carry large kinetic energy and momentum with them. This makes them useful as projectiles: they are used to bombard the nucleus of an atom of one element to convert it into another element, a process called transmutation.

Alpha particles destroy living cells and cause biological damage.

Alpha particles get scattered while passing through thin mica or gold foils — the observation that first revealed the nucleus.

Q3. Alpha particles are stopped by a sheet of paper, yet they are the most damaging of the three radiations once they get inside the body. How can the weakest penetrator be the worst offender?

A. Because the two facts have the same cause. An \(\\alpha\) particle ionises so heavily that it uses up all its energy within a few centimetres — which is why paper stops it outside the body, and why, if the source is taken inside, that whole energy is dumped into living cells packed into a very short path.

12.9 Properties Of Beta Particles

What A Beta Particle Is

Beta particles are fast moving electrons emitted from the nucleus of an atom. A \(\\beta\) particle is represented as \(^{\\ 0}*{-1}\\beta\) or \(^{\\ 0}*{-1}e\).

That raises an obvious difficulty, since a nucleus contains only protons and neutrons, and no electrons at all. The answer is that the electron is not sitting there waiting to leave. Inside the nucleus, a neutron changes into a proton, and this change creates and throws out an electron at the same instant:

\[^{1}*{0}n \\longrightarrow ^{1}*{1}p + ^{\\ 0}\_{-1}e\]

The \(\\beta\) particle does not reside inside the nucleus in the form of an electron; it is created during the emission. The change is drawn in Fig. 12.9-1.

Diagram

Fig. 12.9-1 — Origin Of A Beta Particle A single neutron shown on the left as a plain circle labelled ‘neutron ¹₀n’ inside a dashed outline marked ‘nucleus’. An arrow to the right leads to two objects: a circle marked with a plus sign labelled ‘proton ¹₁p’, which stays inside the dashed nucleus outline, and a small dot labelled ‘electron ⁰₋₁e’ shown flying away out of the nucleus with an arrow labelled ‘emitted as a β particle’.

In the symbol \(^{\\ 0}\_{-1}e\), the upper number is the mass number and the lower number is the charge in units of \(e\). The 0 says a \(\\beta\) particle carries almost no mass compared with a nucleon; the \(-1\) says its charge is one unit negative. Both numbers are counts, not units.

Q4. A nucleus contains only protons and neutrons, and no electrons at all. So where does the electron come from when a nucleus shoots one out as a beta particle?

A. It is made on the spot. A neutron inside the nucleus turns into a proton, and that change creates an electron which is instantly thrown out. Nothing was stored there beforehand.

Mass, Charge And Specific Charge

The rest mass of a \(\\beta\) particle — its mass when it is not moving — is \(9.1 \\times 10^{-31}\\ \\text{kg}\), the same as the mass \(m\_e\) of an electron. Its charge is \(-1.6 \\times 10^{-19}\\ \\text{C}\), or \(-e\).

Its specific charge, the charge carried per kilogram, therefore works out to \(1.76 \\times 10^{11}\\ \\text{C kg}^{-1}\) — several thousand times the value for an \(\\alpha\) particle, since the \(\\beta\) particle is thousands of times lighter.

Beta Particles And Cathode Rays

Both \(\\beta\) particles and cathode rays are fast moving electrons, but they differ in their origin, as Table 12.5 shows.

Table 12.5 — Beta particles compared with cathode rays

Property Beta particles Cathode rays
Nature Fast moving electrons Fast moving electrons
Where they come from The nucleus of the atom The orbital electrons of the atom
Charge and mass Same as those of an electron Same as those of an electron
Behaviour in electric and magnetic fields Deflected Deflected
Speed

The speed of beta particles is of the order of \(10^{8}\\ \\text{m s}^{-1}\), but it is always less than \(3 \\times 10^{8}\\ \\text{m s}^{-1}\), the speed of light, which nothing can reach. A typical value is about 90 per cent of the speed of light, that is nearly \(2.7 \\times 10^{8}\\ \\text{m s}^{-1}\).

Different \(\\beta\) particles emitted from the same radioactive substance have different speeds, because their energies are spread over a range running from zero up to a certain maximum value.

Ionising Power

Beta particles ionise the gas through which they pass, but far less strongly than \(\\alpha\) particles do. Their ionising power is roughly \(1/100\) times that of the \(\\alpha\) particles, but nearly 100 times that of the \(\\gamma\) radiations.

Penetrating Power

Since a \(\\beta\) particle loses its energy more slowly than an \(\\alpha\) particle, it travels further. The penetrating power of \(\\beta\) particles is more than that of \(\\alpha\) particles but less than that of \(\\gamma\) radiations.

Beta particles can travel through nearly 5 m in air and can pass through a thin card sheet, and even through thin aluminium foil. A 5 mm thick aluminium sheet, however, can stop them.

Example 12.9 — Identifying A Radiation By What Stops It

  1. Given: a radiation from a source passes easily through a thick sheet of paper, but is completely stopped by an aluminium sheet 5 mm thick. Asked: name the radiation.

  2. Reasoning used: each radiation has its own stopping material — \(\\alpha\) particles are stopped by thin card or thick paper, \(\\beta\) particles pass through thin card and thin aluminium foil but are stopped by about 5 mm of aluminium, and \(\\gamma\) radiations pass through far more than that.

  3. Since the radiation passed through thick paper, it is not \(\\alpha\). Since 5 mm of aluminium stopped it completely, it is not \(\\gamma\) either, because \(\\gamma\) radiation would go straight through. That leaves only one possibility.

  4. The radiation is \(\\beta\). The check: the two observations bracket it from both sides — too penetrating to be \(\\alpha\), not penetrating enough to be \(\\gamma\).

Behaviour In Electric And Magnetic Fields

Beta particles are negatively charged, so they get deflected by electric and magnetic fields. The deflection of a \(\\beta\) particle is in a direction opposite to that of an \(\\alpha\) particle, since the charge on the \(\\beta\) particle is negative while that on the \(\\alpha\) particle is positive. The deflection of a \(\\beta\) particle is also more than that of an \(\\alpha\) particle, since a \(\\beta\) particle is much lighter.

Other Properties

Beta particles affect a photographic plate.

They cause fluorescence on striking a fluorescent material.

They produce X-rays when they are stopped by metals of high atomic number and high melting point, such as tungsten.

Beta particles cause more biological damage than \(\\alpha\) particles, as they can easily pass through the skin of our body.

The tenth property above is how medical X-rays are made. In an X-ray tube, fast electrons are aimed at a tungsten target; the sudden stop converts part of their energy into X-rays. Tungsten is chosen for the same two reasons given above — a high atomic number, which makes the process efficient, and a very high melting point, so the target survives the heat. The X-rays are then let out through a small window, and the rest of the tube is shielded with lead so that stray radiation does not reach the patient or the operator.

Two radiations, P and Q, are emitted by a radioactive source. P is stopped by a thick sheet of paper, while Q passes through the paper but is stopped by a 5 mm thick sheet of aluminium.

(i) Name the radiations P and Q.

(ii) State which of the two has the greater ionising power.

(iii) State which of the two would be deflected more by a magnetic field, and give a reason.

12.10 Properties Of Gamma Radiations

Nature And Wavelength

Gamma radiations are not particles at all. They are electromagnetic waves, like X-rays and like light, and they differ from those two only in wavelength.

The wavelength of gamma radiations is of the order of \(10^{-4}\\ \\text{nm}\), that is \(10^{-13}\\ \\text{m}\). The wavelength of X-rays lies in the range \(10^{-2}\\ \\text{nm}\) to \(10\\ \\text{nm}\), and that of light in the range \(400\\ \\text{nm}\) to \(800\\ \\text{nm}\). So of the three, gamma radiation has much the shortest wavelength, as Fig. 12.10-1 and Table 12.6 show.

Diagram

Fig. 12.10-1 — Where Gamma Radiations Sit On The Wavelength Scale A single horizontal scale line running left to right, marked with decreasing wavelength towards the left. Mark and label three bands along it: ‘gamma radiations, about 10⁻⁴ nm’ at the far left, ‘X-rays, 10⁻² nm to 10 nm’ in the middle, and ‘light, 400 nm to 800 nm’ at the right. Label the axis ‘wavelength’ and add arrows showing ‘shorter wavelength’ to the left and ‘longer wavelength’ to the right. Draw a short wave squiggle under each band, tightly spaced under gamma and widely spaced under light.

Wavelength is a length, so its SI unit is the metre (m). Very short wavelengths are quoted in nanometre (nm), where \(1\\ \\text{nm} = 10^{-9}\\ \\text{m}\). Converting is a one-step job: to change nm into m, multiply by \(10^{-9}\).

Table 12.6 — Gamma radiations, X-rays and light compared

Property Gamma radiations X-rays Light
Nature Electromagnetic waves Electromagnetic waves Electromagnetic waves
Where they come from The nucleus of the atom Transition of electrons in the inner orbits of an atom Transition of electrons in the outer orbits of an atom
Wavelength About \(10^{-4}\\ \\text{nm}\) \(10^{-2}\\ \\text{nm}\) to \(10\\ \\text{nm}\) \(400\\ \\text{nm}\) to \(800\\ \\text{nm}\)
Speed in vacuum \(3 \\times 10^{8}\\ \\text{m s}^{-1}\) \(3 \\times 10^{8}\\ \\text{m s}^{-1}\) \(3 \\times 10^{8}\\ \\text{m s}^{-1}\)
Effect of electric and magnetic fields Not deflected Not deflected Not deflected

Although X-rays and gamma radiations are similar in their properties, their origin is quite different. X-rays are emitted when there is a transition of electrons in the inner orbits of an atom, whereas gamma radiations are given out from the nucleus.

Example 12.10 — Comparing Two Wavelengths

  1. Given: wavelength of gamma radiation \(\\lambda\_\\gamma = 10^{-4}\\ \\text{nm}\); wavelength of a green light wave \(\\lambda\_L = 500\\ \\text{nm}\). Asked: how many times shorter the gamma wavelength is, and the value of \(\\lambda\_\\gamma\) in metre.

  2. Relations used: \(\\text{ratio} = \\dfrac{\\lambda\_L}{\\lambda\_\\gamma}\), which needs both wavelengths in the same unit — they are already both in nm. For the second part, \(1\\ \\text{nm} = 10^{-9}\\ \\text{m}\).

  3. \(\\text{ratio} = \\dfrac{500\\ \\text{nm}}{10^{-4}\\ \\text{nm}} = 5 \\times 10^{2} \\times 10^{4} = 5 \\times 10^{6}\). And \(\\lambda\_\\gamma = 10^{-4} \\times 10^{-9}\\ \\text{m} = 10^{-13}\\ \\text{m}\).

  4. The gamma wavelength is about \(5 \\times 10^{6}\) times shorter than that of green light, and equals \(10^{-13}\\ \\text{m}\). The ratio is a pure number, since two lengths were divided; the second answer is a length and carries the unit metre, agreeing with the value quoted above.

Speed

The speed of gamma radiations is the same as the speed of light, that is \(3 \\times 10^{8}\\ \\text{m s}^{-1}\) in vacuum or air. This is fixed for all gamma radiations, unlike the speeds of \(\\alpha\) and \(\\beta\) particles, which vary from one emission to the next.

Ionising Power

The ionising power of gamma radiations is very low. It is \(1/10{,}000\) times that of \(\\alpha\) particles and \(1/100\) times that of \(\\beta\) particles — the least of the three.

Penetrating Power

The penetrating power of gamma radiations is high. It is about 10,000 times that of \(\\alpha\) particles and 100 times that of \(\\beta\) particles.

Gamma radiations can travel through 500 m in air and can pass through a 30 cm thick sheet of iron. A thick sheet of lead is required to stop them.

Why The Three Differ So Sharply

Of the three radiations, the gamma radiations are the most energetic, and being uncharged they have a smaller rate of collisions with the atoms or molecules of the medium through which they pass. Fewer collisions means less ionisation but a longer journey — so they have the least ionising power and the maximum penetrating power.

The \(\\alpha\) particles are at the other end. They are the least energetic and therefore suffer more collisions, giving them the maximum ionising power and the least penetrating power. Their energy is soon dissipated.

Other Properties

Like X-rays and light, gamma radiations are not deflected by electric and magnetic fields, since they are not charged particles.

Gamma radiations affect a photographic plate.

They cause fluorescence when they strike a fluorescent material.

They are diffracted by crystals, as X-rays are.

Gamma radiations can easily pass through the human body, and therefore they cause immense biological damage.

Gamma radiations are very useful in the treatment of cancer.

The same penetrating power that makes gamma radiation dangerous also makes it useful for looking inside solid metal. A sealed gamma source is placed on one side of a welded pipe joint and a photographic film on the other; cracks and air pockets inside the weld let more radiation through and show up as dark marks on the developed film, without the joint being cut open. The work is done behind heavy shielding, with the area cleared of people, because a source strong enough to pass through steel will pass through a person just as easily.

A radioactive source gives out all three radiations. The beam is sent through an electric field and then towards a 30 cm thick iron sheet.

(i) Name the radiation that is undeflected by the field, and give a reason.

(ii) Name the radiation that passes through the iron sheet.

(iii) State one property, other than penetrating power, in which this radiation differs from a \(\\beta\) particle.

12.11 Distinction Between The Properties Of Alpha, Beta And Gamma Radiations

The three radiations have now been met one at a time. Set side by side, in Table 12.7, the pattern behind them becomes easier to hold: heaviest and most strongly charged at one end, massless and uncharged at the other, with almost everything else following from that.

Table 12.7 — Properties of alpha, beta and gamma radiations compared

Property \(\\alpha\)-particle \(\\beta\)-particle \(\\gamma\)-radiation
Nature Stream of positively charged particles, that is, helium nuclei Stream of negatively charged particles, that is, energetic electrons Highly energetic electromagnetic radiations
Speed Nearly \(10^{7}\\ \\text{m s}^{-1}\) About 90% of the speed of light, or \(2.7 \\times 10^{8}\\ \\text{m s}^{-1}\) \(3 \\times 10^{8}\\ \\text{m s}^{-1}\) in vacuum
Rest mass Four times the mass of a proton, that is \(6.68 \\times 10^{-27}\\ \\text{kg}\) Equal to the mass of an electron, that is \(9.1 \\times 10^{-31}\\ \\text{kg}\) No mass; rest mass is zero
Charge Positive, two times that of a proton, \(+3.2 \\times 10^{-19}\\ \\text{C}\) (or \(+2e\)) Negative, \(-1.6 \\times 10^{-19}\\ \\text{C}\) (or \(-e\)) No charge
Specific charge (\(q/m\)) \(4.79 \\times 10^{7}\\ \\text{C kg}^{-1}\) \(1.76 \\times 10^{11}\\ \\text{C kg}^{-1}\) —
Wavelength — — \(10^{-4}\\ \\text{nm}\) or \(10^{-13}\\ \\text{m}\)
Effect of electric and magnetic fields Less deflected, and in a direction opposite to that of the \(\\beta\)-particle More deflected than the \(\\alpha\)-particle, in the opposite direction Unaffected
Ionising power Maximum, 10,000 times that of \(\\gamma\) Less than the \(\\alpha\)-particle, 100 times that of \(\\gamma\) Minimum
Penetrating range in air Small, 3 cm to 8 cm Large, up to a few metres Very large, up to a few hundred metres
Stopping substance Thin paper, human skin About 1 mm thick lead, or about 5 mm thick aluminium About 30 cm thick lead sheet, or a few metre thick concrete
Biological damage Less damage More damage Immense damage

Two of the rows carry a dash rather than a value, and the dash is itself information. Specific charge has no meaning for \(\\gamma\) radiation, because there is neither a charge nor a rest mass to divide one by the other. Wavelength has no entry for \(\\alpha\) and \(\\beta\), because those are particles and not waves.

Identifying An Unknown Radiation

In practice a radiation is identified by testing it, not by being told what it is. Two tests are enough, and their order matters: a field test first, since it picks out the uncharged radiation straight away, and a stopping test afterwards. The route is set out in FC 12.1.

Flowchart

FC 12.1 — Identifying A Radiation From Two Tests Radiation from a source → pass it through an electric field → branch A: undeflected → γ radiation → check: passes through 30 cm of iron; branch B: deflected towards the negative plate, smaller deflection → α particle → check: stopped by thick paper; branch C: deflected towards the positive plate, larger deflection → β particle → check: stopped by 5 mm of aluminium

Lead appears twice in the last row for a reason. It is very dense and has a high atomic number, so a large number of heavy atoms are packed into a small thickness, and the radiation meets them within a short distance. That is why lead does in centimetres what concrete needs metres to do.

Three radiations P, Q and R come from a radioactive source. P has zero rest mass, Q has a specific charge of \(1.76 \\times 10^{11}\\ \\text{C kg}^{-1}\), and R has a specific charge of \(4.79 \\times 10^{7}\\ \\text{C kg}^{-1}\).

(i) Name P, Q and R.

(ii) Arrange the three in increasing order of their penetrating power.

(iii) State which of them would be stopped by a sheet of thick paper.

12.12 Changes Within The Nucleus In Alpha, Beta And Gamma Emission

Radioactive substances are not stable. Their nuclei undergo spontaneous decay by emitting either alpha, beta or gamma radiation, and in doing so they form a more stable nucleus. The emission of \(\\alpha\), \(\\beta\) and \(\\gamma\) radiation is a nuclear change.

The three do not change the nucleus in the same way. In the emission of alpha and beta radiation there is a change in the number of protons and neutrons inside the nucleus. In the emission of gamma radiation there is a change only in the energy of the nucleus, and no change at all in its numbers of protons and neutrons.

12.12.1 Alpha Emission

If an unstable nucleus contains many more neutrons than protons, it may throw out a particle made of two protons and two neutrons tightly bound together — the alpha particle. A stream of \(\\alpha\)-particles is called \(\\alpha\)-rays, or \(\\alpha\)-radiation.

Suppose the nucleus X of a radioactive element has mass number \(A\) and atomic number \(Z\), and emits an \(\\alpha\)-particle. A daughter nucleus Y of a new element is formed, whose mass number is \(A - 4\) and whose atomic number is \(Z - 2\). The original nucleus X is called the parent nucleus.

\[^{A}*{Z}\\mathrm{X} \\longrightarrow\\ ^{A-4}*{Z-2}\\mathrm{Y} + ^{4}\_{2}\\mathrm{He}\]

Here \(^{A}*{Z}\\mathrm{X}\) is the parent nucleus, \(^{A-4}*{Z-2}\\mathrm{Y}\) is the daughter nucleus and \(^{4}\_{2}\\mathrm{He}\) is the \(\\alpha\)-particle. Due to the emission of an alpha particle, the atomic number \(Z\) decreases by 2 and the mass number \(A\) decreases by 4.

The resulting nucleus thus has 2 protons and 2 neutrons — 4 nucleons in all — fewer than the original nucleus. The new element has properties similar to those of the element which is two places before it in the periodic table.

X → Y + α A nucleus that fires out an alpha particle loses exactly two protons and two neutrons. So its mass number drops by 4 and its atomic number by 2 — and since the atomic number has changed, the atom is now a different element altogether.

When a radioactive uranium nucleus \(^{238}*{92}\\mathrm{U}\) emits an \(\\alpha\)-particle, a new nucleus, thorium \(^{234}*{90}\\mathrm{Th}\), is formed. The change is shown in Fig. 12.12-1 and written as

\[^{238}*{92}\\mathrm{U} \\longrightarrow\\ ^{234}*{90}\\mathrm{Th} + ^{4}\_{2}\\mathrm{He}\]

Diagram

Fig. 12.12-1 — Alpha Emission By A Uranium Nucleus Three labelled circles in a row with an arrow after the first and a plus sign before the third. First circle, large: ‘92 protons and 146 neutrons’, labelled ²³⁸₉₂U, captioned ‘parent nucleus’. Arrow. Second circle, slightly smaller: ‘90 protons and 144 neutrons’, labelled ²³⁴₉₀Th, captioned ‘daughter nucleus’. Plus sign. Third circle, small: ‘2 protons and 2 neutrons’, labelled ⁴₂He, captioned ‘alpha particle’.

Counting across that equation is worth doing once. On the left there are 92 protons and 146 neutrons. On the right there are \(90 + 2 = 92\) protons and \(144 + 2 = 146\) neutrons. Nothing has been created or lost, so both the atomic number and the mass number are conserved, exactly as a nuclear reaction requires.

Example 12.11 — Writing The Equation For An Alpha Emission

  1. Given: a radium nucleus \(^{226}*{88}\\mathrm{Ra}\) emits an \(\\alpha\)-particle. Asked: the mass number and atomic number of the daughter nucleus, and the equation for the change.

  2. Relations used: in \(\\alpha\)-emission the mass number decreases by 4 and the atomic number decreases by 2, because the \(\\alpha\)-particle carries away 2 protons and 2 neutrons.

  3. Mass number of the daughter \(= 226 - 4 = 222\). Atomic number of the daughter \(= 88 - 2 = 86\). The element of atomic number 86 is radon, Rn, so the change is

\[^{226}*{88}\\mathrm{Ra} \\longrightarrow\\ ^{222}*{86}\\mathrm{Rn} + ^{4}*{2}\\mathrm{He}\]

  1. The daughter nucleus is \(^{222}\_{86}\\mathrm{Rn}\), with mass number 222 and atomic number 86. These are counts and carry no unit. The check: \(222 + 4 = 226\) and \(86 + 2 = 88\), so both numbers balance across the equation.

12.12.2 Beta Emission

When an unstable nucleus contains more neutrons than protons, a neutron may change into a proton by emitting an electron, which keeps the charge conserved. The electron given out from the nucleus at a high speed is the beta particle.

\[^{1}*{0}n \\longrightarrow\\ ^{1}*{1}p + ^{\\ 0}\_{-1}e\]

A stream of \(\\beta\)-particles is called \(\\beta\)-rays.

Because a nuclear reaction has to obey certain conservation laws, another particle called the anti-neutrino, written \(\\bar{\\upsilon}\), was also taken to be emitted along with the \(\\beta\)-particle. This particle is uncharged and its rest mass is almost zero, so it changes neither the charge nor the mass numbers. The complete transformation equation is

\[^{1}*{0}n \\longrightarrow\\ ^{1}*{1}p + ^{\\ 0}\_{-1}e + \\bar{\\upsilon}\]

So in emitting a \(\\beta\)-particle, the number of nucleons in the nucleus stays the same, but the number of neutrons decreases by one while the number of protons increases by one. In other words, the mass number \(A\) does not change, while the atomic number \(Z\) increases by one.

The nucleus therefore becomes a nucleus of a new element, with properties similar to those of the element one place higher in the periodic table. Since the daughter has the same mass number as the parent but a different atomic number, the daughter product is an isobar of the parent atom.

If a radioactive nucleus P with mass number \(A\) and atomic number \(Z\) emits a beta particle to form a daughter nucleus Q with mass number \(A\) and atomic number \(Z + 1\), the change is written as

\[^{A}*{Z}\\mathrm{P} \\longrightarrow\\ ^{A}*{Z+1}\\mathrm{Q} + ^{\\ 0}\_{-1}e\]

A radioactive carbon nucleus \(^{14}*{6}\\mathrm{C}\), having 14 nucleons — 6 protons and 8 neutrons — emits a \(\\beta\)-particle and changes to a new nucleus, nitrogen \(^{14}*{7}\\mathrm{N}\), having 14 nucleons, this time 7 protons and 7 neutrons:

\[^{14}*{6}\\mathrm{C} \\longrightarrow\\ ^{14}*{7}\\mathrm{N} + ^{\\ 0}\_{-1}e\]

In \(\\alpha\)-emission both numbers fall: \(A\) by 4 and \(Z\) by 2. In \(\\beta\)-emission the mass number does not move at all and \(Z\) rises by 1. The one distinction: an \(\\alpha\)-particle takes nucleons away, while a \(\\beta\)-particle only converts one nucleon into another kind.

12.12.3 Gamma Emission

In many cases an alpha or a beta emission is found to be followed by a \(\\gamma\)-emission. It occurs when the daughter or the parent nucleus is left in a state of excitation, that is, holding an excess of energy.

This extra energy is released in the form of electromagnetic radiation known as \(\\gamma\)-radiation, or a \(\\gamma\)-ray photon. The \(\\gamma\)-ray takes no mass and no electric charge away from the nucleus. No neutrons or protons are lost, and so the nucleus does not decay into a different nucleus: there is no change in the mass number \(A\) and no change in the atomic number \(Z\) in gamma emission.

\[^{A}*{Z}\\mathrm{X}^{\*} \\longrightarrow\\ ^{A}*{Z}\\mathrm{X} + \\gamma\]

Here the star indicates the excited state of the nucleus. In gamma emission the excited nucleus simply comes down to its ground state, as drawn in Fig. 12.12-2.

Diagram

Fig. 12.12-2 — Gamma Emission From An Excited Nucleus Two horizontal energy levels drawn one above the other, the upper one labelled ‘excited state, ᴬZX*’ and the lower one labelled ‘ground state, ᴬZX’. A downward arrow joins the upper level to the lower one, and a wavy arrow leaves the nucleus sideways, labelled ‘γ-ray photon carries away the extra energy’. Beside each level draw a small identical circle marked ‘same Z, same A’ to show that the composition of the nucleus does not change.

Q. Does a nucleus become a nucleus of a different element when it emits a gamma ray?

A. No. Its atomic number is unchanged, so it is the same element — only with less energy than before.

In a single radioactive decay, \(\\alpha\) and \(\\beta\) particles are never emitted simultaneously. There will be either an \(\\alpha\)-emission or a \(\\beta\)-emission, and either of them may be accompanied by a \(\\gamma\)-emission.

Decay Continues Till A Stable Nucleus Is Formed

The daughter product formed after the emission of either an \(\\alpha\)-particle or a \(\\beta\)-particle may itself be radioactive, and it may decay further by emitting an \(\\alpha\)-particle or a \(\\beta\)-particle of its own. The process continues, step after step, till a stable nucleus is formed, as shown in FC 12.2.

Flowchart

FC 12.2 — Why A Decay Continues In Steps Unstable parent nucleus → emits an α or a β particle → daughter nucleus formed → is this daughter stable? → branch A: yes → decay stops here; branch B: no → this daughter now acts as the parent → emits an α or a β particle → next daughter nucleus formed → repeat the test until a stable nucleus is reached

Example 12.12 — Two Steps Of A Decay Chain

  1. Given: a thorium nucleus \(^{232}*{90}\\mathrm{Th}\) first emits an \(\\alpha\)-particle, and the daughter so formed then emits a \(\\beta\)-particle. Asked: the mass number and atomic number of the nucleus formed at each step.

  2. Relations used: \(\\alpha\)-emission lowers \(A\) by 4 and \(Z\) by 2; \(\\beta\)-emission leaves \(A\) unchanged and raises \(Z\) by 1. The steps must be taken in order, since the second acts on the product of the first.

  3. After the \(\\alpha\)-emission: \(A = 232 - 4 = 228\) and \(Z = 90 - 2 = 88\), which is radium, \(^{228}*{88}\\mathrm{Ra}\). After the \(\\beta\)-emission from this nucleus: \(A = 228\) and \(Z = 88 + 1 = 89\), which is actinium, \(^{228}*{89}\\mathrm{Ac}\).

\[^{232}*{90}\\mathrm{Th} \\longrightarrow\\ ^{228}*{88}\\mathrm{Ra} + ^{4}*{2}\\mathrm{He}, \\qquad ^{228}*{88}\\mathrm{Ra} \\longrightarrow\\ ^{228}*{89}\\mathrm{Ac} + ^{\\ 0}*{-1}e\]

  1. The first step gives \(^{228}*{88}\\mathrm{Ra}\) and the second gives \(^{228}\_{89}\\mathrm{Ac}\). Both numbers balance in each equation, and the final nucleus is an isobar of the one before it, as a \(\\beta\)-emission requires.

12.12.4 Effect On Atomic Number And Mass Number

Everything in this section reduces to six entries, collected in Table 12.8. They are worth knowing outright, because every decay equation is built from them.

Table 12.8 — Effect on atomic number (Z) and mass number (A) due to alpha, beta and gamma emission

Quantity \(\\alpha\) emission \(\\beta\) emission \(\\gamma\) emission
Atomic number \(Z\) decreases by 2 increases by 1 no change
Mass number \(A\) decreases by 4 no change no change

A nucleus \(^{A}\_{Z}\\mathrm{X}\) emits, one after another, an \(\\alpha\)-particle, then a \(\\beta\)-particle, then a \(\\gamma\)-ray.

(i) Write the mass number and the atomic number of the nucleus formed after each of the three steps, in terms of \(A\) and \(Z\).

(ii) State which of the three steps leaves the element unchanged, and give a reason.

(iii) State the relationship between the nucleus formed after the second step and the nucleus that existed before it.

12.13 Uses Of Radioactivity — Radio Isotopes

Radioactivity is not confined to the heavy elements beyond lead. The isotopes of some elements with atomic number \(Z < 82\) are also found to be radioactive, and these are called the radio isotopes.

The nucleus of an atom becomes radioactive when the number of neutrons in it exceeds the number of protons inside it. This is the situation with the isotopes of many otherwise stable elements. For example, cobalt \(^{60}*{27}\\mathrm{Co}\) with \(Z = 27\) and \(A = 60\), carbon \(^{14}*{6}\\mathrm{C}\) with \(Z = 6\) and \(A = 14\), potassium \(^{40}*{19}\\mathrm{K}\) with \(Z = 19\) and \(A = 40\), and phosphorus \(^{32}*{15}\\mathrm{P}\) with \(Z = 15\) and \(A = 32\), are all radio isotopes.

Radio isotopes are also prepared artificially by nuclear transmutation, because they find vital use in three fields: medical, scientific and industrial.

12.13.1 Medical Uses

Many diseases such as leukemia and cancer are cured by radio therapy. Gamma radiations from cobalt-60, \(^{60}\_{27}\\mathrm{Co}\), are used to treat cancer by killing the cells in the malignant tumour of the patient. For radio therapy the radiation must be able to penetrate the human skin, which rules out the radiations that cannot.

The salts of weak radioactive isotopes, such as radio-sodium chloride, radio-iron and radio-iodine, are used for diagnosis. Such radio isotopes are called tracers, because they can be followed from outside the body. Tracers are used to detect suspected brain tumours and blood clots before they become dangerous.

Tracers are also used to study the natural processes going on in the human body. For example, radio-sodium chloride is injected along with common sodium chloride, and the radioactivity of the blood at different parts of the body is then tested to study the blood circulation. This process is called radio cardiology.

Gamma rays emitted by radio isotopes are used to sterilize bandages, dressings, syringes and other equipment, making them free of germs. This method is quicker, more reliable and cheaper than the process of sterilization by heat.

A tracer and a therapy source are used for opposite reasons. A tracer is a weak source, meant only to be detected and to leave the tissue unharmed. A therapy source such as cobalt-60 is a strong one, chosen precisely because its radiation kills cells. One is for finding out, the other for treating.

Example 12.13 — Choosing The Right Radiation For A Job

  1. Given: a tumour lying deep inside a patient’s body is to be treated with radiation, and three sources are available — one emitting \(\\alpha\) particles, one emitting \(\\beta\) particles and one emitting \(\\gamma\) radiations. Asked: which source should be used, and why.

  2. Reasoning used: the radiation must first reach the tumour, so its penetrating power decides the choice. Alpha particles are stopped by the human skin, beta particles are stopped by a few millimetres of aluminium and travel only a short way through tissue, while gamma radiations pass easily through the human body.

  3. An \(\\alpha\) source is ruled out at once, since its radiation would not get past the skin. A \(\\beta\) source would not reach a tumour lying deep inside. Only the \(\\gamma\) source emits radiation able to penetrate that far.

  4. The \(\\gamma\) source is the right choice, which is why cobalt-60 is used in radio therapy. The check: the same property that makes \(\\gamma\) radiation the most dangerous to a healthy body is the property that lets it be aimed at a tumour deep inside one.

12.13.2 Scientific Uses

Alpha particles emitted from radio isotopes are used as projectiles for nuclear reactions. The scattering of alpha particles from the nucleus helps us in estimating the size of the nucleus and in understanding the nature of nuclear forces.

Radioactive tracers are used in agricultural science to study the growth of plants that have been given a particular chemical manure. In this way it can be found out how readily a plant takes in phosphate, and to which part of the plant the phosphate goes.

The age of excavated material of archaeological importance, of rocks, and hence of buried plants, is estimated by studying the rate of decay of carbon-14 by \(\\beta\) emissions in the remains of dead plants:

\[^{14}*{6}\\mathrm{C} \\longrightarrow\\ ^{14}*{7}\\mathrm{N} + ^{\\ 0}\_{-1}e\]

This process is called carbon dating.

Carbon dating works only on what was once alive. A living plant keeps taking in carbon, including its radioactive isotope, from its surroundings. Once it dies the taking in stops, the carbon-14 already inside goes on decaying, and how little of it remains is a measure of how long ago the plant died.

12.13.3 Industrial Uses

Radio isotopes such as \(^{235}\_{92}\\mathrm{U}\) are used as fuel for the atomic energy reactors.

Radio isotopes are used by engineers in factories to avoid the accumulation of charge on the moving parts of machines due to friction. The radiations emitted by the isotope ionise the air around the machine, and the ions so produced carry away with them the charge accumulated on the moving parts.

The ionising effect of the radiations from radio isotopes is also used in making certain luminescent signs.

Knowing the penetrating power of the \(\\beta\)-radiations emitted from radio isotopes, they are used to control the thickness of paper, plastic and metal sheets during their manufacture. The method works because a thicker sheet absorbs more of the beta radiation than a thinner one does, so the amount getting through is a direct measure of the thickness. The arrangement is a continuous loop, shown in FC 12.3.

Flowchart

FC 12.3 — Controlling Sheet Thickness With Beta Radiation Beta source placed above the moving sheet → radiation passes through the sheet → detector below the sheet counts what gets through → count compared with the value set for the correct thickness → branch A: count too low, so the sheet is too thick → rollers pressed closer together; branch B: count too high, so the sheet is too thin → rollers eased apart; branch C: count correct → rollers left as they are → the sheet moves on and the detector reads again, so the loop repeats continuously

A factory uses a radioactive source and a detector to check the thickness of aluminium foil moving on rollers.

(i) State, with a reason, whether an \(\\alpha\) source or a \(\\beta\) source is suitable for this job.

(ii) Explain what the detector reading would do if the foil suddenly became thinner.

(iii) State why a \(\\gamma\) source would be a poor choice for measuring the thickness of a thin foil.

12.14 Sources Of Harmful Radiations

The main sources of harmful radiations are three: radioactive fall out from the nuclear power plants, nuclear waste, and other sources such as cosmic radiation and X-rays.

Radioactive Fall Out From The Nuclear Power Plants

Nuclear power plants are now a major source of electricity in the world. In normal working they are shielded, and the radiations stay inside. The danger arises if something goes wrong.

If somehow there is an accident in the reactor of a power plant, a large amount of radioactive material and radiations will escape out into the atmosphere. This will affect not only the population living around the plant, but also life at far off places, because air currents can carry the radioactive material a long way from where it started. That journey is set out in FC 12.4.

Flowchart

FC 12.4 — How Radioactive Fall Out Spreads After A Reactor Accident Accident inside the reactor → radioactive material and radiations escape from the damaged reactor building → material enters the atmosphere as fine dust and gas → air currents carry it away from the plant → material settles on land and water, both nearby and at far off places → it enters soil, crops, drinking water and living bodies → people near the plant and people far from it are both affected

Three such accidents have already occurred — in the U.S.A. on 28 March 1979, in Ukraine on 26 April 1986, and in Japan on 11 March 2011. These accidents caused damage to the reactor buildings and injuries to the persons working in the plants, as well as to persons in their vicinity.

Nuclear Waste

The fuel rods used in the nuclear power reactors are rejected when their activity decreases below a certain level. They then become nuclear waste.

As these rods are still quite radioactive and are a source of harmful radiations, they should not be dumped in open garbage. They can contaminate water and soil, and so affect human beings and other living organisms. How such waste is to be dealt with is taken up later in this chapter.

Fall out and nuclear waste are different problems. Fall out is radioactive material that escapes by accident and is carried about by the air. Nuclear waste is the spent fuel produced in normal working, which is known about in advance and must be stored safely rather than released.

Cosmic Radiation, X-rays And Other Sources

The cosmic radiations enter the earth’s atmosphere from the sun and from outer space. They contain a large number of high energy particles.

Most of the charged particles among them are deviated by the earth’s magnetic field, so they never reach us. But a substantial amount of uncharged radiations, such as \(\\gamma\)-rays and X-rays, is not turned aside by a magnetic field and does reach the earth’s atmosphere. Some of these radiations are quite harmful for human beings. The situation is drawn in Fig. 12.14-1.

Diagram

Fig. 12.14-1 — Cosmic Radiation Reaching The Earth The earth drawn as a sphere at the lower right, with a thin shaded band above its surface labelled ‘atmosphere’ and curved magnetic field lines drawn around it labelled ‘earth’s magnetic field’. From the upper left, arrows labelled ‘cosmic radiation from the sun and outer space’ come in towards the earth. Show two kinds: arrows labelled ‘charged particles’ curving away along the field lines with the note ‘deviated by the magnetic field’, and straight arrows labelled ‘γ-rays and X-rays, uncharged’ passing through the field lines undeviated and reaching the atmosphere.

Apart from this, the high energy particle accelerators used in research also produce harmful X-rays.

Cosmic radiation is thinned out by the air above us, so there is more of it the higher you go. This is why an airline pilot or a member of a cabin crew, flying for many hours a year at a height of about 10 km, receives a measurably larger dose over a year than someone working at ground level.

(i) Name the two main sources of harmful radiation associated with a nuclear power plant, and state which of the two occurs only if something goes wrong.

(ii) Explain why most of the charged particles in cosmic radiation never reach the earth’s surface, while \(\\gamma\)-rays from space do.

(iii) State one reason why spent fuel rods must not be dumped in open garbage.

12.15 Harmful Effects Of Radiation

The radiations produced in a nuclear reaction include alpha radiations, beta radiations, neutrons, gamma rays and X-rays. When they meet living tissue they interact with it within about \(10^{-14}\\ \\text{s}\) and cause biological damage.

The damage is done by ionisation. A radiation passing through a cell knocks electrons out of the molecules it meets, and molecules that have been broken up in this way can no longer do the job the cell needs them to do.

Three Kinds Of Biological Effect

The harmful biological effects of nuclear radiations are of three types, set out in Table 12.9.

Table 12.9 — The three types of harmful biological effect

Feature Short term effects Long term effects Genetic effects
Examples Diarrhoea, sore throat, loss of hair, nausea Leukemia and cancer Genes in the living cell get modified, or mutated
Can they be recovered from? Yes, they are recoverable No, they are irrecoverable Not applicable; the change is passed on
Who is affected Only the individual actually exposed Only the individual actually exposed The later generations of the person exposed

The third row is the one worth pausing over. The first two kinds of effect are limited to the individuals who are actually exposed to the radiations, while the third type appears in the later generations of the person exposed, because the radiation has altered the genes rather than the person.

Loss of hair, an upset stomach and a fall in the number of blood cells appear early in radiation sickness for a common reason. These tissues make new cells fastest, and a cell is most easily damaged by radiation at the moment it is dividing.

Acute And Chronic Exposure

Exposure to radiation can be acute or chronic. It is acute if there is an accidental burst of radiations from an unshielded source — a large dose received in a short time. It is chronic in the case of persons working in atomic energy establishments, who may get exposed to small amounts of such radiations over long periods.

Acute and chronic describe how the dose arrives, not how serious it is. Acute means a large dose all at once, from an accident with an unshielded source. Chronic means small amounts received again and again over months or years, as in the course of work.

Effects Of X-rays And Gamma Rays

The exposure of a pregnant woman to X-rays for a long time may affect the child in the womb, and raises the risk of cancer and of leukemia in the child. This is why an X-ray examination is never carried out casually, and why a woman is asked whether she may be pregnant before one is taken. The exposure of the upper part of the body to X-rays may damage the thyroid glands and can cause cancer, which is the reason a lead collar is placed over the throat during a dental X-ray.

Exposure to \(\\gamma\)-rays destroys the cells of the body. It affects the blood cells, the gastro-intestinal tract, and the reproductive and hair cells, and it can harm the DNA and RNA of living cells as well.

The uncharged neutrons produced in nuclear reactions also cause biological damage. Being uncharged, they are not stopped by the materials that stop \(\\alpha\) and \(\\beta\) particles, which is why shielding against them has to be designed separately.

Every one of these effects is a reason for the precautions taken in the next section rather than a reason for alarm. Radiation is dangerous in the way that mains electricity is dangerous: harmful when handled carelessly, and safe when the known precautions are followed.

(i) Name the type of harmful effect of radiation that appears not in the person exposed but in later generations, and state what the radiation has altered in order to produce it.

(ii) Distinguish between acute exposure and chronic exposure, stating the basis of the difference.

(iii) Give a reason why a lead collar is placed over a patient’s throat during a dental X-ray.

12.16 Safety Precautions While Using Nuclear Energy

Electricity these days is also generated by the use of nuclear energy obtained from nuclear power plants. In a nuclear power plant, a radioactive source such as uranium-235 or plutonium-239 is used, and a controlled chain fission reaction is carried out in it.

Countries differ on how large a part nuclear power should play. Those in favour point to the very large amount of electricity obtained from a small quantity of fuel, and without the smoke of a coal plant; those against point to the cost of building the plant safely and to waste that stays radioactive for a very long time. What is not in dispute is that wherever nuclear energy is used, three aspects require special care: the establishment of nuclear power plants, the handling of radioactive materials, and the safe disposal of nuclear waste.

12.16.1 Establishing A Nuclear Power Plant

While establishing a nuclear power plant to generate electricity, the following precautions must be taken.

It must be ensured that the people working in the plant are not exposed to nuclear radiations, and that in case of any accident there is a minimum spread of radiations.

The nuclear reactor of the power plant must be shielded with lead and steel walls, so as to stop radiations from escaping out to the environment during its normal operation.

The nuclear reactor must be housed in an airtight building of strong concrete structure, which can withstand earthquakes, fires and explosions. Airtight matters as much as strong: it is what keeps radioactive gas and dust from getting out even if something inside is damaged.

There must be a back-up cooling system for the reactor core, so that in case of failure of one system the other cooling system could take its place, and the core is saved from over heating and melting.

12.16.2 Handling Radioactive Materials

People working with radioactive materials are required to follow strictly the safety measures given below.

They should put on special lead lined aprons and lead gloves.

They should handle the radioactive materials with long lead tongs.

The safety limit for each type of radiation is known, therefore care must be taken to ensure that no one is exposed beyond the safety limit in any case. For this, special film badges are used, which are tested from time to time to know the amount of radiations to which a particular person has been exposed.

The radioactive substance must be kept in a thick lead container with a very narrow opening, as shown in Fig. 12.16-1. The walls of the container absorb the radiations which strike them from the inside, and thus radiations only come out through the narrow opening.

Diagram

Fig. 12.16-1 — Storing And Handling A Radioactive Source A thick-walled container drawn in cross-section, the walls hatched and labelled ‘thick lead walls’, with a small radioactive source at the centre labelled ‘radioactive substance’. Draw several arrows from the source striking the inner walls and stopping there, labelled ‘radiations absorbed by the walls’, and one arrow escaping upward through a small gap at the top labelled ‘narrow opening’. To one side show a gloved hand holding a pair of long tongs reaching towards the container, labelled ‘long lead tongs, lead gloves and lead lined apron’.

How thick the shielding needs to be depends on which radiation is being stopped. Alpha particles can easily be stopped by a thin metal sheet, but for \(\\beta\)-particles a thick metal sheet is needed, whereas for gamma radiations very thick lead sheets are required.

The tongs are long for a reason that has nothing to do with lead. Radiation spreads out from a source in all directions, so the further away you stand, the smaller the share of it that reaches you. Distance is free shielding, and it is the easiest precaution of all to take.

A film badge is a small piece of photographic film sealed inside a light-proof holder, clipped to the clothing of anyone working with radiation — in a hospital X-ray room, a research laboratory or a power plant. Radiation darkens the film in exactly the way it darkened Becquerel’s plate. At the end of each period the film is developed, and how dark it has become gives the dose that person received. Different filters in the holder let the reading be split up between the different radiations, so that both the size and the kind of exposure are known.

12.16.3 Safe Disposal Of Nuclear Waste

The radioactive material left after its use is known as nuclear waste. The nuclear waste obtained from laboratories, hospitals, scientific establishments or power plants must first be kept in thick casks, and then it must be buried in specially constructed deep underground stores. These stores must be made quite far from the populated areas.

The casks can also be buried in useless mines, and these mines must be sealed after storing the casks. The route the waste takes is shown in FC 12.5.

Flowchart

FC 12.5 — Safe Disposal Of Nuclear Waste Used radioactive material from a laboratory, hospital, scientific establishment or power plant → sealed inside thick casks → casks taken to a site far from populated areas → branch A: buried in specially constructed deep underground stores; branch B: buried in disused mines → mine or store sealed after the casks are placed in it → site kept under watch so that the waste stays contained

(i) State why a radioactive source is kept in a thick lead container with only a narrow opening, rather than in a completely sealed one.

(ii) Give a reason why a back-up cooling system is required for a reactor core.

(iii) State two conditions that a site chosen for burying nuclear waste must satisfy.

12.17 Background Radiations

Put a radiation detector down in an ordinary room, with no radioactive source anywhere near it, and it will still register counts. What it is picking up is the background radiation.

NoteDefinition

Background Radiations The background radiations are the radioactive radiations — such as \(\\alpha\), \(\\beta\) and \(\\gamma\) — to which we all are exposed, even in the absence of a visible radioactive source.

The Two Types Of Source

There are two types of sources of background radiations, the internal and the external, as set out in Table 12.10 and shown in Fig. 12.17-1.

Table 12.10 — Internal and external sources of background radiation

Feature Internal source External source
Where it is Inside our own body Outside the body, in our surroundings
What it consists of Radioactive substances present inside the body, such as potassium (K-40), carbon (C-14) and radium Cosmic rays, naturally occurring radioactive elements such as radon-222, and solar radiations
Part of the body affected K-40 and C-14 are present in the entire body and affect all parts equally; radium is largely found in the bones, so it affects only the bones and the reproductive organs The whole body, from outside

Diagram

Fig. 12.17-1 — Sources Of Background Radiation A simple outline figure of a person standing on the ground. Inside the outline place three small labels with short arrows pointing outwards from within the body: ‘potassium K-40’, ‘carbon C-14’ and ‘radium, mainly in the bones’, grouped under the heading ‘internal sources’. Outside the figure draw arrows coming towards the person: from above, two arrows labelled ‘cosmic rays’ and ‘solar radiations’; from the ground below, an arrow labelled ‘radon-222 from rocks and soil’. Group these under the heading ‘external sources’.

Radon-222 is a gas, and it seeps out of the rocks and soil on which houses stand. Out in the open it disperses at once, but indoors it can collect, which is why buildings in some regions are deliberately designed with good under-floor ventilation.

Why We Need Not Worry About It

It is not possible for us to keep ourselves away from the background radiations. They come from the ground we stand on, the sky above us and our own bodies.

But the total background radiation from both the internal and the external sources does not exceed the maximum permissible dose laid down for human safety. Within that limit there is nothing to worry about.

The effect of different radiations on human tissue is measured as a dose, in the unit sievert (Sv). Smaller doses are given in millisievert (mSv), where \(1\\ \\text{mSv} = 10^{-3}\\ \\text{Sv}\). One chest X-ray gives a patient about \(0.2\\ \\text{mSv}\).

Example 12.14 — Adding Up A Dose

  1. Given: one chest X-ray gives a dose of \(0.2\\ \\text{mSv}\). Asked: the total dose received from three such X-rays taken in a year, expressed in mSv and in Sv.

  2. Relation used: \(\\text{total dose} = \\text{dose per exposure} \\times \\text{number of exposures}\), since doses received one after another simply add up. For the second part, \(1\\ \\text{mSv} = 10^{-3}\\ \\text{Sv}\).

  3. \(\\text{total dose} = 0.2\\ \\text{mSv} \\times 3 = 0.6\\ \\text{mSv}\), and \(0.6\\ \\text{mSv} = 0.6 \\times 10^{-3}\\ \\text{Sv} = 6 \\times 10^{-4}\\ \\text{Sv}\).

  4. The total dose is \(0.6\\ \\text{mSv}\), that is \(6 \\times 10^{-4}\\ \\text{Sv}\). The unit is retained at every step, and the answer is sensible: three exposures give three times the dose of one.

(i) Name two radioactive substances present inside the human body that contribute to background radiation.

(ii) State one external source of background radiation, and say where it comes from.

(iii) A dental X-ray gives a dose of \(0.005\\ \\text{mSv}\). Calculate the total dose, in mSv, received from four such X-rays.

12.18 Nuclear Energy And Mass–Energy Equivalence

In a nuclear change due to a radioactive phenomenon such as decay, fission or fusion, the total sum of the masses of the product nuclei is always less than the total sum of the masses of the reactant nuclei. Something has gone missing. This is called the loss in nuclear mass.

In 1905, on the basis of his theory of relativity, Einstein suggested that mass and energy are interchangeable — that mass is one of the forms energy can take. The missing mass has therefore not vanished; it has been released as energy.

NotePrinciple / Law

Einstein’s Mass–Energy Equivalence Mass and energy are interchangeable. If a nuclear change involves a loss in mass \(\\Delta m\), the energy \(E\) released is given by \[E = (\\Delta m)c^{2}\] where \(\\Delta m\) is the loss in mass in kilogram (kg), \(c\) is the speed of light, equal to \(3 \\times 10^{8}\\ \\text{m s}^{-1}\), and \(E\) is the energy in joule (J). The energy so obtained is called nuclear energy.

E = Δm × c² The mass that goes missing in a nuclear change reappears as energy. Because it is multiplied by the square of the speed of light, which is an enormous number, even a tiny loss of mass gives out a very large amount of energy.

Q5. Splitting a heavy nucleus releases energy, and joining two light nuclei also releases energy. What single fact about mass could possibly explain both?

A. In both cases the products together weigh less than what went in. That lost mass is what comes out as energy, following \(E = (\\Delta m)c^{2}\) — so it does not matter whether the nucleus was split or joined, only that mass was lost.

12.18.1 Energy Equivalent To A Mass Of One Kilogram

Take the simplest case and put \(\\Delta m = 1\\ \\text{kg}\) into the relation. When 1 kg of mass is lost, the amount of energy released is

\[E = (\\Delta m)c^{2} = 1 \\times (3 \\times 10^{8})^{2} = 9 \\times 10^{16}\\ \\text{J}\]

This is often more meaningful in the unit in which household electricity is sold. Since \(1\\ \\text{kWh} = 3.6 \\times 10^{6}\\ \\text{J}\),

\[E = \\frac{9 \\times 10^{16}}{3.6 \\times 10^{6}}\\ \\text{kWh} = 2.5 \\times 10^{10}\\ \\text{kWh}\]

Thus 1 kg of mass is equivalent to \(9 \\times 10^{16}\\ \\text{J}\), or \(2.5 \\times 10^{10}\\ \\text{kWh}\), of energy. No ordinary process comes anywhere near converting a whole kilogram, but the figure shows why even a fractional loss of mass matters so much.

Example 12.15 — Energy From A Small Loss Of Mass

  1. Given: loss in mass \(\\Delta m = 2\\ \\text{g}\); speed of light \(c = 3 \\times 10^{8}\\ \\text{m s}^{-1}\). Asked: the energy released, in joule and in kWh.

  2. Relation used: \(E = (\\Delta m)c^{2}\). The relation requires \(\\Delta m\) in kilogram, so the mass must be converted first: \(2\\ \\text{g} = 2 \\times 10^{-3}\\ \\text{kg}\).

  3. \(E = (2 \\times 10^{-3}\\ \\text{kg}) \\times (3 \\times 10^{8}\\ \\text{m s}^{-1})^{2} = 2 \\times 10^{-3} \\times 9 \\times 10^{16}\\ \\text{J} = 1.8 \\times 10^{14}\\ \\text{J}\). In kWh, \(E = \\dfrac{1.8 \\times 10^{14}}{3.6 \\times 10^{6}}\\ \\text{kWh} = 5 \\times 10^{7}\\ \\text{kWh}\).

  4. The energy released is \(1.8 \\times 10^{14}\\ \\text{J}\), that is \(5 \\times 10^{7}\\ \\text{kWh}\). The check: 2 g is \(2 \\times 10^{-3}\) times 1 kg, and \(2 \\times 10^{-3} \\times 2.5 \\times 10^{10}\\ \\text{kWh}\) does give \(5 \\times 10^{7}\\ \\text{kWh}\), agreeing with the figure worked out above.

12.18.2 Energy Equivalent To One Atomic Mass Unit

A kilogram is far too large a unit for a single nucleus. The mass of atomic particles is expressed instead in the atomic mass unit (a.m.u.), also called the unified unit (u), where

\[1\\ \\text{a.m.u.} = \\frac{1}{12} \\times \\text{mass of one atom of carbon-12}\]

Since \(6.02 \\times 10^{26}\) atoms of carbon-12 have a mass of 12 kg,

\[1\\ \\text{a.m.u.} = \\frac{1}{12} \\times \\frac{12\\ \\text{kg}}{6.02 \\times 10^{26}} = 1.66 \\times 10^{-27}\\ \\text{kg}\]

Now put this into the mass–energy relation. Due to a loss in mass \(\\Delta m = 1\\ \\text{a.m.u.}\), the energy released is

\[E = (1.66 \\times 10^{-27}) \\times (3 \\times 10^{8})^{2}\\ \\text{J} = 1.49 \\times 10^{-10}\\ \\text{J}\]

Nuclear energies are quoted in MeV rather than in joule, and \(1\\ \\text{MeV} = 1.6 \\times 10^{-13}\\ \\text{J}\). So

\[E = \\frac{1.49 \\times 10^{-10}}{1.6 \\times 10^{-13}}\\ \\text{MeV} = 931\\ \\text{MeV}\]

\[1\\ \\text{a.m.u.} = 931\\ \\text{MeV}\]

Thus 1 a.m.u. of mass is equivalent to 931 MeV of energy. If in a nuclear reaction 1 a.m.u. of mass is lost, then 931 MeV of energy is released. This single figure turns a mass loss straight into an energy, and it is used constantly in the two sections that follow.

Three units are now in play. Mass: \(1\\ \\text{a.m.u.} = 1.66 \\times 10^{-27}\\ \\text{kg}\). Energy: \(1\\ \\text{MeV} = 1.6 \\times 10^{-13}\\ \\text{J}\). And the bridge between them: \(1\\ \\text{a.m.u.} = 931\\ \\text{MeV}\). Check which unit the question wants before substituting.

The masses of the three particles making up an atom are given in all three of these units in Table 12.11.

Table 12.11 — Mass of the electron, proton and neutron in kg, a.m.u. and MeV

Particle Mass in kg Mass in a.m.u. Mass in MeV
Electron \(9.1 \\times 10^{-31}\) 0.00055 0.511
Proton \(1.6726 \\times 10^{-27}\) 1.00727 938.3
Neutron \(1.6749 \\times 10^{-27}\) 1.00865 939

Example 12.16 — From A Mass Loss In A.M.U. To An Energy In MeV

  1. Given: in a nuclear reaction the loss in mass is \(\\Delta m = 0.03\\ \\text{a.m.u.}\) Asked: the energy released, in MeV.

  2. Relation used: \(1\\ \\text{a.m.u.} = 931\\ \\text{MeV}\), so the energy released \(=\) the loss in mass in a.m.u. multiplied by \(931\\ \\text{MeV}\). Working in these units avoids converting to kilogram and back again.

  3. \(E = 0.03 \\times 931\\ \\text{MeV} = 27.93\\ \\text{MeV}\)

  4. The energy released is \(27.93\\ \\text{MeV}\). The size is sensible for a nuclear reaction, where energies run to a few MeV or a few hundred MeV, and not to the few eV of a chemical change.

(i) In a nuclear reaction the loss in mass is \(0.25\\ \\text{a.m.u.}\) Calculate the energy released, in MeV.

(ii) Calculate the energy released, in joule, when a mass of \(0.1\\ \\text{g}\) is lost, taking \(c = 3 \\times 10^{8}\\ \\text{m s}^{-1}\).

(iii) Express the answer to part (ii) in kWh, given that \(1\\ \\text{kWh} = 3.6 \\times 10^{6}\\ \\text{J}\).

12.19 Nuclear Fission

12.19.1 The Fission Of A Uranium-235 Nucleus

NoteDefinition

Nuclear Fission Nuclear fission is the process in which a heavy nucleus splits into two lighter nuclei of nearly the same size, when bombarded with slow neutrons. In each fission reaction, a tremendous amount of energy, nearly \(190\\ \\text{MeV}\), is released.

The reason for the release of energy is the one established in the last section. The sum of the masses of the product nuclei is less than the sum of the mass of the parent nucleus and the mass of the neutron, so there is a loss of mass in this reaction, and this loss in mass is converted into energy according to Einstein’s mass-energy relation \(E = (\\Delta m)c^{2}\).

In 1939, Otto Hahn and Fritz Strassmann found that when a slow neutron strikes the \(^{235}*{92}\\mathrm{U}\) nucleus, it gets absorbed in it and a most unstable isotope \(^{236}*{92}\\mathrm{U}\) is formed, which immediately splits into two nuclei — barium \(^{144}*{56}\\mathrm{Ba}\) and krypton \(^{89}*{36}\\mathrm{Kr}\) — with the release of three neutrons and a tremendous amount of energy. The splitting was explained shortly afterwards by Lise Meitner and Otto Frisch, who gave the process its name. It is shown in Fig. 12.19-1 and expressed as

\[^{235}*{92}\\mathrm{U} + ^{1}*{0}n \\longrightarrow \\left(^{236}*{92}\\mathrm{U}\\right) \\longrightarrow\\ ^{144}*{56}\\mathrm{Ba} + ^{89}*{36}\\mathrm{Kr} + 3,^{1}*{0}n + \\text{energy}\]

Diagram

Fig. 12.19-1 — Fission Of A Uranium-235 Nucleus Left to right. A small dot labelled ‘slow neutron ¹₀n’ moving with an arrow towards a large circle labelled ²³⁵₉₂U. Next, a dashed circle labelled ‘(²³⁶₉₂U) highly unstable’ shown vibrating. Next, the dashed circle splitting into two unequal circles flying apart along arrows: the larger labelled ¹⁴⁴₅₆Ba and the smaller labelled ⁸⁹₃₆Kr. From the point of splitting draw three small dots flying outwards labelled ‘3 neutrons’, and a wavy arrow labelled ‘energy released’.

Later on, it was found that the highly unstable isotope \(^{236}*{92}\\mathrm{U}\) decays not only into \(^{144}*{56}\\mathrm{Ba}\) and \(^{89}*{36}\\mathrm{Kr}\) fragments, but that it can decay into certain other pairs of nuclei as well, such as \(\\left(^{141}*{56}\\mathrm{Ba},\\ ^{92}*{36}\\mathrm{Kr}\\right)\) and \(\\left(^{148}*{57}\\mathrm{La},\\ ^{85}\_{35}\\mathrm{Br}\\right)\), for which the nuclear reactions are

\[^{235}*{92}\\mathrm{U} + ^{1}*{0}n \\longrightarrow \\left(^{236}*{92}\\mathrm{U}\\right) \\longrightarrow\\ ^{141}*{56}\\mathrm{Ba} + ^{92}*{36}\\mathrm{Kr} + 3,^{1}*{0}n + \\text{energy}\]

\[^{235}*{92}\\mathrm{U} + ^{1}*{0}n \\longrightarrow \\left(^{236}*{92}\\mathrm{U}\\right) \\longrightarrow\\ ^{148}*{57}\\mathrm{La} + ^{85}*{35}\\mathrm{Br} + 3,^{1}*{0}n + \\text{energy}\]

In each fission reaction, the atomic number \(Z\) and the mass number \(A\) remain conserved. Checking the first of these equations: on the left \(A = 235 + 1 = 236\), and on the right \(A = 141 + 92 + 3 = 236\); on the left \(Z = 92\), and on the right \(Z = 56 + 36 = 92\), since a neutron carries no charge.

The two fragments obtained on the fission of one U-235 nucleus are themselves unstable, and they emit \(\\beta\)-particles and \(\\gamma\) rays to become stable.

Example 12.17 — Completing A Fission Equation

  1. Given: the fission reaction \(^{235}*{92}\\mathrm{U} + ^{1}*{0}n \\longrightarrow\\ ^{140}*{54}\\mathrm{Xe} + ^{94}*{38}\\mathrm{Sr} + x,^{1}\_{0}n + \\text{energy}\). Asked: the number \(x\) of neutrons released.

  2. Relation used: in a nuclear reaction the mass number is conserved, so the total \(A\) on the left equals the total \(A\) on the right. Each neutron contributes 1 to the mass number.

  3. Total \(A\) on the left \(= 235 + 1 = 236\). Total \(A\) of the two nuclei on the right \(= 140 + 94 = 234\). So the neutrons must carry \(236 - 234 = 2\), giving \(x = 2\).

  4. Two neutrons are released. The check is on the atomic number, which must balance too: on the left \(Z = 92\), and on the right \(Z = 54 + 38 = 92\), with the neutrons adding nothing, so the equation is consistent.

12.19.2 Energy Released In One Fission Reaction

Due to the fission of one uranium nucleus, nearly 190 MeV of energy is released. Where that figure comes from is worth following through, because it is the whole argument of this chapter in one calculation.

Take the fission reaction

\[^{235}*{92}\\mathrm{U} + ^{1}*{0}n \\longrightarrow\\ ^{144}*{56}\\mathrm{Ba} + ^{89}*{36}\\mathrm{Kr} + 3,^{1}\_{0}n + \\text{energy}\]

and take the masses as: neutron \(= 1.01\\ \\text{a.m.u.}\), uranium-235 nucleus \(= 234.99\\ \\text{a.m.u.}\), barium-144 nucleus \(= 143.87\\ \\text{a.m.u.}\), krypton-89 nucleus \(= 88.90\\ \\text{a.m.u.}\)

The loss in mass in the fission of one nucleus is the mass that went in minus the mass that came out:

\[\\Delta m = \\left(234.99 + 1.01\\right) - \\left(143.87 + 88.90 + 3 \\times 1.01\\right)\\ \\text{a.m.u.}\]

\[\\Delta m = 236.00 - 235.80 = 0.20\\ \\text{a.m.u.}\]

From the mass-energy equivalence, and using \(1\\ \\text{a.m.u.} = 931\\ \\text{MeV}\),

\[\\text{energy released} = 0.20 \\times 931\\ \\text{MeV} = 186.2\\ \\text{MeV}\]

which is the nearly 190 MeV quoted above. Notice how small the mass loss is — 0.20 out of 236, less than a tenth of one per cent — and how large the energy it gives.

The major part of this energy is obtained in the form of the kinetic energy of the fragments obtained from the fission, and the remaining part in the form of the kinetic energy of the neutrons emitted, \(\\gamma\)-rays, heat and light.

Example 12.18 — The Energy Of One Fission In Joule

  1. Given: energy released in one fission of a U-235 nucleus \(= 186.2\\ \\text{MeV}\); \(1\\ \\text{MeV} = 1.6 \\times 10^{-13}\\ \\text{J}\). Asked: this energy in joule.

  2. Relation used: \(\\text{energy in J} = \\text{energy in MeV} \\times 1.6 \\times 10^{-13}\), since each MeV is worth that many joule.

  3. \(E = 186.2 \\times 1.6 \\times 10^{-13}\\ \\text{J} = 297.92 \\times 10^{-13}\\ \\text{J} = 2.98 \\times 10^{-11}\\ \\text{J}\)

  4. The energy released in one fission is about \(2.98 \\times 10^{-11}\\ \\text{J}\). The answer is sensible: a single nucleus is a very small thing, and the enormous energy of a reactor comes from an enormous number of such fissions rather than from any one of them.

Due to the fission of 1 g of \(^{235}\_{92}\\mathrm{U}\), which contains about \(2.56 \\times 10^{21}\) nuclei, nearly \(4.8 \\times 10^{23}\\ \\text{MeV}\) of energy is released, from which about \(2.1 \\times 10^{4}\\ \\text{kWh}\) of electrical energy can be obtained. That much energy is liberated in the explosion of nearly 20 tonnes of trinitrotoluene (TNT).

The ore of uranium found in nature contains two isotopes — 99.3% of \(^{238}*{92}\\mathrm{U}\) and 0.7% of \(^{235}*{92}\\mathrm{U}\). Although both the isotopes are fissionable, it is found experimentally that the isotope \(^{235}*{92}\\mathrm{U}\) is more easily fissionable than the isotope \(^{238}*{92}\\mathrm{U}\). The reason is that the fission of a \(^{238}*{92}\\mathrm{U}\) nucleus is possible only by fast neutrons, while the fission of a \(^{235}*{92}\\mathrm{U}\) nucleus can occur even by slow neutrons.

12.19.3 Controlled And Uncontrolled Chain Reactions

Each fission of a uranium-235 nucleus releases three new neutrons. Those neutrons do not simply fly away: under suitable conditions they can cause fission in other uranium nuclei nearby.

So a chain of fission of nuclei is formed which, once started, continues till the entire uranium is consumed, and the energy obtained from the nuclear fission continuously increases. The branching is drawn in Fig. 12.19-2.

Diagram

Fig. 12.19-2 — A Chain Reaction A branching tree drawn left to right. A single dot labelled ‘neutron’ strikes a circle labelled ²³⁵₉₂U; from it emerge two fragment circles and three dots labelled ‘neutrons’. Each of those three neutrons strikes another ²³⁵₉₂U circle, and each of those releases its own two fragments and three neutrons, making nine. Show the nine going on to a further row drawn faintly. Label the rows ‘first fission’, ‘second stage: 3 fissions’, ‘third stage: 9 fissions’ and add the note ‘each stage releases more energy than the one before’.

Due to such a chain reaction, a strong explosion of the entire uranium occurs in a very short interval of time, and a tremendous amount of energy is released which can be very harmful. This is the uncontrolled chain reaction, and a nuclear bomb is based on it.

If the chain reaction is controlled, the same energy can be put to use. It is controlled by absorbing some of the neutrons emitted in the fission process by means of cadmium rods, and then making the remaining neutrons slow by the moderators, such as graphite or heavy water. In this way the rate of energy generated in fission can be controlled and utilised for constructive purposes. This is the principle of a nuclear reactor, set out in FC 12.6.

Flowchart

FC 12.6 — Keeping A Chain Reaction Under Control In A Reactor Slow neutron absorbed by a ²³⁵₉₂U nucleus → nucleus undergoes fission → two fragments plus three fast neutrons plus energy released → cadmium rods absorb the surplus neutrons → moderator of graphite or heavy water slows down the neutrons that remain → on average one slow neutron goes on to cause the next fission → the reaction continues at a steady rate → heat carried away and used to generate electricity → cycle repeats from the first step

Both chain reactions are the same physics; only the number of neutrons allowed to continue differs. In an uncontrolled chain reaction every neutron goes on to cause a fresh fission, so the rate grows explosively. In a controlled one, cadmium rods absorb the surplus so that on average just one neutron continues, and the rate stays steady.

Stand a row of dominoes so that the first one topples two, each of those topples two more, and so on in a branching pattern. Tip the first one over and watch how quickly the numbers grow. Then take a few dominoes out of the middle rows and try again — you have just modelled what the cadmium rods do.

12.19.4 Uses Of Nuclear Fission

The tremendous amount of energy released in the process of fission can be used in two ways: for the destructive use, and for the constructive use.

For the destructive use, the fission process is used in a nuclear bomb — sometimes also called an atom bomb, though that word is a misnomer, since the energy comes from the nucleus and not from the atom as a whole. Here the energy released is fast and uncontrolled.

For the constructive use, the fission process is used in a nuclear reactor, where the rate of release of energy is slow and controlled. This energy is used to generate electric power.

(i) In the reaction \(^{235}*{92}\\mathrm{U} + ^{1}*{0}n \\longrightarrow\\ ^{137}*{52}\\mathrm{Te} + ^{97}*{40}\\mathrm{Zr} + x,^{1}\_{0}n + \\text{energy}\), find the value of \(x\), showing how you used the conservation of mass number.

(ii) In a fission reaction the loss in mass is \(0.18\\ \\text{a.m.u.}\) Calculate the energy released, in MeV.

(iii) Name the substance used to absorb the surplus neutrons in a nuclear reactor, and state what would happen if it were removed.

12.20 Distinction Between Radioactive Decay And Nuclear Fission

Radioactive decay and nuclear fission are both nuclear phenomena in which a nucleus disintegrates. That much they share. Everything else about them differs, as Table 12.12 shows.

Table 12.12 — Radioactive decay compared with nuclear fission

Feature Radioactive decay Nuclear fission
How it starts It is a spontaneous process; the nucleus decays by itself It does not occur by itself; it is initiated when neutrons are bombarded on a heavy nucleus
What happens to the nucleus, and how much energy is released The nucleus emits either an \(\\alpha\) or a \(\\beta\) particle, with the emission of energy in the form of \(\\gamma\) rays, which is not very large The nucleus splits into two nearly equal fragments, and a tremendous amount of energy is released
Can the rate be controlled? The rate of radioactive decay cannot be controlled The rate of nuclear fission can be controlled

Q. Why can the rate of fission be controlled while the rate of decay cannot?

A. Because fission needs neutrons to keep it going, and those can be absorbed. Decay needs nothing from outside, so there is nothing to take away.

A sample of uranium-235 is kept in a laboratory. Some of its nuclei decay by emitting \(\\alpha\) particles, and the sample is later used as fuel in a reactor, where its nuclei undergo fission.

(i) State which of the two processes, decay or fission, needs something from outside in order to begin.

(ii) Give a reason why the energy released per nucleus is far greater in the second process than in the first.

12.21 Nuclear Fusion

12.21.1 Examples Of Fusion Reactions

Fission split a heavy nucleus into lighter ones. Fusion does the opposite.

NoteDefinition

Nuclear Fusion Nuclear fusion is the process in which two light nuclei combine to form a heavy nucleus. In this process also, a huge amount of energy is released.

The reason for the release of energy is the same as before. The mass of the product nucleus is less than the sum of the masses of the two combining nuclei, and this loss in mass is released in the form of energy according to the mass-energy equivalence relation \(E = (\\Delta m)c^{2}\).

Take deuterium, the isotope \(^{2}*{1}\\mathrm{H}\). When two deuterium nuclei fuse, \(3.3\\ \\text{MeV}\) of energy is released and the nucleus of the helium isotope \(^{3}*{2}\\mathrm{He}\) is formed. This helium isotope then fuses with one more deuterium nucleus to form a helium nucleus \(^{4}\_{2}\\mathrm{He}\), and \(18.3\\ \\text{MeV}\) of energy is released in this second step. The nuclear reactions are

\[^{2}*{1}\\mathrm{H} + ^{2}*{1}\\mathrm{H} \\longrightarrow\\ ^{3}*{2}\\mathrm{He} + ^{1}*{0}n + 3.3\\ \\text{MeV}\]

\[^{3}*{2}\\mathrm{He} + ^{2}*{1}\\mathrm{H} \\longrightarrow\\ ^{4}*{2}\\mathrm{He} + ^{1}*{1}\\mathrm{H} + 18.3\\ \\text{MeV}\]

Thus, in all, three deuterium nuclei fuse to form a helium nucleus with a release of \(21.6\\ \\text{MeV}\) of energy. A part of this energy is obtained in the form of the kinetic energy of the neutron \(^{1}*{0}n\) and of the proton \(^{1}*{1}\\mathrm{H}\). The two steps are collected in FC 12.7.

Flowchart

FC 12.7 — Three Deuterium Nuclei Forming One Helium Nucleus Deuterium ²₁H + deuterium ²₁H → helium isotope ³₂He + neutron ¹₀n + 3.3 MeV → this ³₂He now fuses with a third deuterium ²₁H → helium ⁴₂He + proton ¹₁H + 18.3 MeV → net result: three deuterium nuclei used, one helium nucleus formed, 21.6 MeV of energy released in all

Diagram

Fig. 12.21-1 — Fusion Of Two Deuterium Nuclei Two small circles approaching each other along arrows from the left and the right, each marked ‘1p, 1n’ and labelled ²₁H, with a plus sign shown on each to indicate positive charge. At the centre they merge into a single circle marked ‘2p, 1n’ and labelled ³₂He, with a small dot flying off labelled ‘neutron ¹₀n’ and a wavy arrow labelled ‘energy released, 3.3 MeV’.

Other examples of fusion reactions are

\[^{1}*{1}\\mathrm{H} + ^{1}*{1}\\mathrm{H} \\longrightarrow\\ ^{2}*{1}\\mathrm{H} + ^{\\ 0}*{+1}e + 0.42\\ \\text{MeV}\]

\[^{2}*{1}\\mathrm{H} + ^{2}*{1}\\mathrm{H} \\longrightarrow\\ ^{3}*{1}\\mathrm{H} + ^{1}*{1}\\mathrm{H} + 4.0\\ \\text{MeV}\]

\[^{2}*{1}\\mathrm{H} + ^{3}*{1}\\mathrm{H} \\longrightarrow\\ ^{4}*{2}\\mathrm{He} + ^{1}*{0}n + 17.6\\ \\text{MeV}\]

In the first of these, two protons form a deuterium nucleus; the particle \(^{\\ 0}\_{+1}e\) written alongside is a positron, which carries away the surplus positive charge and keeps the atomic number balanced. In the second, two deuterium nuclei form tritium and a proton, and in the third, deuterium and tritium together form helium and a neutron.

Example 12.19 — Identifying The Product Of A Fusion Reaction

  1. Given: the fusion reaction \(^{2}*{1}\\mathrm{H} + ^{3}*{1}\\mathrm{H} \\longrightarrow\\ ^{A}*{Z}\\mathrm{X} + ^{1}*{0}n + \\text{energy}\). Asked: the mass number, the atomic number and the name of the nucleus X.

  2. Relations used: in a nuclear reaction both the mass number and the atomic number are conserved, so each must balance across the equation.

  3. Mass number: \(2 + 3 = A + 1\), so \(A = 4\). Atomic number: \(1 + 1 = Z + 0\), since the neutron carries no charge, so \(Z = 2\). The element of atomic number 2 is helium.

  4. The nucleus X is \(^{4}\_{2}\\mathrm{He}\), a helium nucleus. The check: two light nuclei have combined to give a heavier one, which is what fusion means.

Example 12.20 — Energy Released In A Fusion Reaction

  1. Given: in the reaction \(^{2}*{1}\\mathrm{H} + ^{3}*{1}\\mathrm{H} \\longrightarrow\\ ^{4}*{2}\\mathrm{He} + ^{1}*{0}n\), the loss in mass is \(\\Delta m = 0.0189\\ \\text{a.m.u.}\) Asked: the energy released, in MeV.

  2. Relation used: \(1\\ \\text{a.m.u.} = 931\\ \\text{MeV}\), so the energy released is the loss in mass in a.m.u. multiplied by \(931\\ \\text{MeV}\).

  3. \(E = 0.0189 \\times 931\\ \\text{MeV} = 17.6\\ \\text{MeV}\)

  4. The energy released is \(17.6\\ \\text{MeV}\), which is the value quoted for this reaction above — so the mass loss and the energy quoted agree with each other.

12.21.2 Why Fusion Needs A Very High Temperature

Nuclear fusion is not possible at ordinary temperature and ordinary pressure. The reason lies in the charge on a nucleus.

When two nuclei approach each other, then due to their positive charge the electrostatic force of repulsion between them becomes so strong that they do not fuse. The closer they get, the harder they are pushed apart, so at ordinary speeds they simply turn away from each other, as Fig. 12.21-2 shows.

Diagram

Fig. 12.21-2 — Repulsion Between Two Approaching Nuclei Two panels. Left panel: two circles each marked with a plus sign, moving slowly towards each other along short arrows, with outward-pointing force arrows between them labelled ‘electrostatic force of repulsion’; show them turning away before touching, labelled ‘at ordinary temperature, no fusion’. Right panel: the same two circles moving towards each other along long arrows labelled ‘high kinetic energy at about 10⁷ K’; show them reaching each other and merging, labelled ‘repulsion overcome, nuclei fuse’.

Hence to make fusion possible, a high temperature of about \(10^{7}\\ \\text{K}\) and a high pressure are required. At such a high temperature, both nuclei, due to their thermal agitations, acquire sufficient kinetic energy to overcome the force of repulsion between them when they approach each other, and so they get fused. This is why the fusion reaction is also called the thermo-nuclear reaction.

Take two bar magnets and slowly push their like poles towards each other. Feel how the push-back grows stronger the closer they come, and how hard you must shove to get them to touch. Two nuclei behave the same way, and the shove that gets them together is supplied by heat.

The Hydrogen Bomb

The working of the hydrogen bomb is based on nuclear fusion. The high temperature of about \(10^{7}\\ \\text{K}\) required for the fusion reaction is produced by the prior explosion of a nuclear fission bomb, which acts only as a way of reaching that temperature.

12.21.3 Fusion In The Sun And The Stars

The source of energy of the sun and the stars is the nuclear fusion of light nuclei, such as hydrogen, present in their core at a very high temperature of about \(10^{7}\\ \\text{K}\) and a high pressure. The thermo-nuclear reaction going on there converts hydrogen into helium nuclei with the release of a tremendous amount of energy:

\[^{1}*{1}\\mathrm{H} + ^{1}*{1}\\mathrm{H} + ^{1}*{1}\\mathrm{H} + ^{1}*{1}\\mathrm{H} \\longrightarrow\\ ^{4}*{2}\\mathrm{He} + 2,^{\\ 0}*{+1}\\beta + 2\\upsilon + \\text{energy}\\ (26.7\\ \\text{MeV})\]

Here the four \(^{1}*{1}\\mathrm{H}\) are four hydrogen nuclei, \(^{4}*{2}\\mathrm{He}\) is the helium nucleus formed, \(^{\\ 0}\_{+1}\\beta\) is a positron and \(\\upsilon\) is a neutrino. Thus in this process, four hydrogen nuclei fuse to form a helium nucleus with a release of \(26.7\\ \\text{MeV}\) of energy.

Why Fusion Is Worth Pursuing On Earth

Although it appears that the energy released in nuclear fusion is much less than that released in nuclear fission — 26.7 MeV against 190 MeV — that comparison is misleading. It compares one reaction with one reaction, not one kilogram with one kilogram.

For a given mass, the energy released due to the fusion of light nuclei is much more than the energy released due to the fission of the same mass of a heavy nucleus. The reason is that for the same mass, the number of light nuclei is much greater than the number of heavy nuclei. Hence the energy obtained per unit mass in fusion is much larger than the energy obtained per unit mass in fission.

This is why scientists are now working hard to get energy by the process of fusion of deuterium, which can be obtained in abundance from sea water at a very low cost. Here the main difficulty is that at \(10^{7}\\ \\text{K}\) the substance gets ionised — it is in the plasma state — and it is very difficult to store. Nowadays, efforts are being made to preserve the substance in the plasma state with the help of nano carbon tubes.

Plasma is often called the fourth state of matter. Heat a gas far enough and its atoms lose their electrons, leaving a mixture of free electrons and bare nuclei. Being charged, plasma responds to electric and magnetic fields — which is exactly what makes it possible to hold it away from the walls of a container.

(i) In the reaction \(^{2}*{1}\\mathrm{H} + ^{2}*{1}\\mathrm{H} \\longrightarrow\\ ^{A}*{Z}\\mathrm{X} + ^{1}*{1}\\mathrm{H} + \\text{energy}\), find \(A\) and \(Z\) and name the nucleus X.

(ii) In a fusion reaction the loss in mass is \(0.0045\\ \\text{a.m.u.}\) Calculate the energy released, in MeV.

(iii) State why this reaction cannot be carried out at ordinary temperature, and give the approximate temperature that would be needed.

12.22 Distinction Between Nuclear Fission And Nuclear Fusion

Fission and fusion are opposite processes — one splits a heavy nucleus, the other joins two light ones — and they differ in almost every practical respect, as Table 12.13 sets out.

Table 12.13 — Nuclear fission compared with nuclear fusion

Feature Nuclear fission Nuclear fusion
What happens Neutrons are bombarded on a heavy nucleus and it splits into two nearly equal light fragments At a very high temperature and high pressure, two light nuclei combine to form a heavy nucleus
Conditions needed Possible at ordinary temperature and ordinary pressure Possible only at a very high temperature, about \(10^{7}\\ \\text{K}\), and a very high pressure
Energy released in one reaction Nearly \(190\\ \\text{MeV}\) Nearly \(24.7\\ \\text{MeV}\)
Energy released for the same mass Less than that in the fusion process Much more than that in the fission process
Is the substance radioactive? The fissionable substance is radioactive, so it gives out harmful radiations and creates a problem in the disposal of its waste The fusionable substance is not radioactive, so it does not give out harmful radiations, and disposal of its waste is also not difficult
Availability Only a limited amount of fissionable substance is available in nature The fusionable substance is found in abundance
Can it be controlled? The fission process can be controlled; a nuclear reactor is based on the controlled fission reaction The fusion reaction cannot be controlled, which is why a fusion reactor could not be constructed so far
Weapon based on it The nuclear bomb is based on the uncontrolled fission reaction The hydrogen bomb is based on the uncontrolled fusion reaction

The figure of about \(24.7\\ \\text{MeV}\) in the third row is the energy obtained from the loss in mass of the nuclei taking part in the fusion reaction that goes on in the sun. Placed beside the \(190\\ \\text{MeV}\) of a single fission, it looks small — but the fourth row is the row that matters for a power supply, since what a fuel is judged by is the energy it gives per kilogram, not per reaction.

In both the processes, nuclear fission and nuclear fusion, the similarity is the release of energy due to a loss in mass.

Q. Which of the two leaves the smaller waste problem, and why?

A. Fusion. Its fuel and products are not radioactive, whereas the fissionable substance is, and its spent fuel must be stored safely for a long time.

(i) Name the process, fission or fusion, that releases more energy per unit mass of fuel, and give the reason for it.

(ii) State two reasons why a fusion reactor has not yet been built, although a fission reactor has.

(iii) State one similarity between nuclear fission and nuclear fusion.

12.23 Summary

  1. An atom consists of electrons, protons and neutrons. The protons and neutrons lie in the nucleus at the centre, while the electrons revolve around it in stationary orbits, in which they do not radiate out energy.

  2. The maximum number of electrons a shell can hold is \(2n^{2}\), where \(n\) is the number of the shell. For \(n = 1\) to 7 the shells are named K, L, M, N, O, P and Q, holding at most 2, 8, 18, 32, 50, 72 and 98 electrons. The energy of an electron increases as \(n\) increases.

  3. The size of an atom is of the order of \(10^{-10}\\ \\text{m}\), while the size of its nucleus is of the order of \(10^{-15}\\ \\text{m}\) to \(10^{-14}\\ \\text{m}\), that is \(10^{-5}\) to \(10^{-4}\) times the size of the atom.

  4. The electron has charge \(-1.6 \\times 10^{-19}\\ \\text{C}\) and mass \(9.1 \\times 10^{-31}\\ \\text{kg}\); the proton has charge \(+1.6 \\times 10^{-19}\\ \\text{C}\) and mass \(1.67 \\times 10^{-27}\\ \\text{kg}\); the neutron has zero charge and mass \(1.67 \\times 10^{-27}\\ \\text{kg}\).

  5. Protons and neutrons together are called nucleons. A nucleus containing \(Z\) protons carries a charge \(+Ze\), and since electrons are so light, the mass of an atom can be taken as the mass of its nucleus.

  6. The atomic number \(Z\) is the number of protons in the nucleus, and the mass number \(A\) is the total number of nucleons. Hence \[\\text{number of neutrons} = A - Z\] where \(A\) and \(Z\) are counts and carry no unit.

  7. In a neutral atom, number of electrons \(=\) number of protons \(= Z\). An atom is written as \(^{A}\_{Z}\\mathrm{X}\), where X is the chemical symbol of the element.

  8. A chemical change alters only the orbital electrons and involves energies of a few eV. A nuclear change alters the nucleons inside the nucleus and involves energies of a few MeV, nearly \(10^{6}\) times as much.

  9. In every nuclear reaction, the atomic number \(Z\) and the mass number \(A\) are conserved: their totals are the same before and after.

  10. Isotopes are atoms of the same element having the same \(Z\) but different \(A\). They have the same chemical properties, since they have the same number of electrons.

  11. Isobars are atoms of different elements having the same \(A\) but different \(Z\), and therefore different numbers of neutrons and electrons.

  12. Isotones are atoms having the same number of neutrons \(A - Z\), but different \(Z\) and different \(A\).

  13. A nucleus is radioactive when its number of neutrons much exceeds its number of protons. The isotopes of nearly all elements with \(Z\) higher than 82 are naturally radioactive.

  14. Radioactivity was discovered by Henry Becquerel in 1896, when a uranium salt wrapped in black paper affected a photographic plate kept in the dark. The radiations were later found to be of three kinds — positively charged \(\\alpha\), negatively charged \(\\beta\) and uncharged \(\\gamma\).

  15. Radioactivity is a nuclear phenomenon: the process of spontaneous emission of \(\\alpha\) or \(\\beta\) and \(\\gamma\) radiations from the nucleus of atoms during their decay. It is unaffected by any physical or chemical change, and the decay of an individual nucleus is random and cannot be predicted.

  16. In an electric or a magnetic field the three radiations separate: \(\\alpha\) particles bend towards the negative plate, \(\\beta\) particles bend the opposite way and more, being much lighter, and \(\\gamma\) radiations pass undeviated because they are uncharged.

  17. An \(\\alpha\) particle is a helium nucleus \(^{4}\_{2}\\mathrm{He}\), of mass \(6.68 \\times 10^{-27}\\ \\text{kg}\) and charge \(+3.2 \\times 10^{-19}\\ \\text{C}\), with specific charge \(4.79 \\times 10^{7}\\ \\text{C kg}^{-1}\) and speed of the order of \(10^{7}\\ \\text{m s}^{-1}\). It has the maximum ionising power and the least penetrating power, being stopped by thick paper after 3 cm to 8 cm of air.

  18. A \(\\beta\) particle is a fast moving electron \(^{\\ 0}\_{-1}e\) created in the nucleus when a neutron changes into a proton, of rest mass \(9.1 \\times 10^{-31}\\ \\text{kg}\) and charge \(-1.6 \\times 10^{-19}\\ \\text{C}\), with specific charge \(1.76 \\times 10^{11}\\ \\text{C kg}^{-1}\) and speed of the order of \(10^{8}\\ \\text{m s}^{-1}\). It travels nearly 5 m in air and is stopped by 5 mm of aluminium.

  19. A \(\\gamma\) radiation is an electromagnetic wave of wavelength about \(10^{-4}\\ \\text{nm}\), that is \(10^{-13}\\ \\text{m}\), travelling at \(3 \\times 10^{8}\\ \\text{m s}^{-1}\), with zero rest mass and no charge. It travels up to a few hundred metres in air and needs about 30 cm of lead to stop it. X-rays come from the inner orbits of an atom, while \(\\gamma\) rays come from the nucleus.

  20. Ionising power is in the order \(\\alpha > \\beta > \\gamma\), and penetrating power in the order \(\\gamma > \\beta > \\alpha\). Biological damage from an external source is least for \(\\alpha\) and greatest for \(\\gamma\).

  21. In alpha emission the mass number falls by 4 and the atomic number by 2: \[^{A}*{Z}\\mathrm{X} \\longrightarrow\\ ^{A-4}*{Z-2}\\mathrm{Y} + ^{4}*{2}\\mathrm{He}\] for example \(^{238}*{92}\\mathrm{U} \\longrightarrow\\ ^{234}*{90}\\mathrm{Th} + ^{4}*{2}\\mathrm{He}\).

  22. In beta emission a neutron changes into a proton, \(^{1}*{0}n \\longrightarrow\\ ^{1}*{1}p + ^{\\ 0}*{-1}e\), so the mass number is unchanged and the atomic number rises by 1: \[^{A}*{Z}\\mathrm{P} \\longrightarrow\\ ^{A}*{Z+1}\\mathrm{Q} + ^{\\ 0}*{-1}e\] for example \(^{14}*{6}\\mathrm{C} \\longrightarrow\\ ^{14}*{7}\\mathrm{N} + ^{\\ 0}\_{-1}e\). The daughter is an isobar of the parent.

  23. In gamma emission an excited nucleus releases its excess energy without losing any nucleon, so neither \(A\) nor \(Z\) changes: \[^{A}*{Z}\\mathrm{X}^{\*} \\longrightarrow\\ ^{A}*{Z}\\mathrm{X} + \\gamma\]

  24. Effect on the two numbers: \(\\alpha\) emission lowers \(Z\) by 2 and \(A\) by 4; \(\\beta\) emission raises \(Z\) by 1 and leaves \(A\) unchanged; \(\\gamma\) emission changes neither. A daughter product may itself be radioactive, and decay continues until a stable nucleus is formed.

  25. Radio isotopes are radioactive isotopes of elements with \(Z < 82\), such as \(^{60}*{27}\\mathrm{Co}\), \(^{14}*{6}\\mathrm{C}\), \(^{40}*{19}\\mathrm{K}\) and \(^{32}*{15}\\mathrm{P}\). They are used medically in radio therapy, as tracers for diagnosis and in sterilizing equipment; scientifically as projectiles, as tracers in agriculture, and in carbon dating through \[^{14}*{6}\\mathrm{C} \\longrightarrow\\ ^{14}*{7}\\mathrm{N} + ^{\\ 0}\_{-1}e\] and industrially as reactor fuel, to remove static charge, in luminescent signs and to control the thickness of sheets.

  26. The main sources of harmful radiation are radioactive fall out from nuclear power plants, nuclear waste, and other sources such as cosmic radiation and X-rays from accelerators.

  27. The harmful biological effects of radiation are of three types: short term recoverable effects, long term irrecoverable effects such as leukemia and cancer, and genetic effects which appear in later generations. Exposure is acute when a large dose arrives at once, and chronic when small doses are received over long periods.

  28. Safety requires shielding the reactor with lead and steel inside a strong airtight concrete building with a back-up cooling system; lead aprons, lead gloves, long tongs, film badges and thick lead containers with narrow openings for those handling sources; and disposal of nuclear waste in thick casks buried in sealed deep underground stores far from populated areas.

  29. Background radiations are those to which we are all exposed even without any visible source. Their internal sources are K-40, C-14 and radium within the body, and their external sources are cosmic rays, radon-222 and solar radiations. The total stays within the maximum permissible dose.

  30. Mass and energy are interchangeable. For a loss in mass \(\\Delta m\), \[E = (\\Delta m)c^{2}\] where \(\\Delta m\) is in kilogram (kg), \(c = 3 \\times 10^{8}\\ \\text{m s}^{-1}\) and \(E\) is in joule (J). Hence a mass of 1 kg is equivalent to \(9 \\times 10^{16}\\ \\text{J}\), or \(2.5 \\times 10^{10}\\ \\text{kWh}\).

  31. \(1\\ \\text{a.m.u.} = 1.66 \\times 10^{-27}\\ \\text{kg}\), and using \(1\\ \\text{MeV} = 1.6 \\times 10^{-13}\\ \\text{J}\), \[1\\ \\text{a.m.u.} = 931\\ \\text{MeV}\]

  32. Nuclear fission is the splitting of a heavy nucleus into two lighter nuclei of nearly the same size when bombarded with slow neutrons, for example \[^{235}*{92}\\mathrm{U} + ^{1}*{0}n \\longrightarrow\\ ^{144}*{56}\\mathrm{Ba} + ^{89}*{36}\\mathrm{Kr} + 3,^{1}\_{0}n + \\text{energy}\] A loss in mass of \(0.20\\ \\text{a.m.u.}\) gives \(186.2\\ \\text{MeV}\), nearly 190 MeV, per fission.

  33. The neutrons released can cause further fissions, giving a chain reaction. Uncontrolled, it is the basis of the nuclear bomb; controlled by cadmium rods, which absorb surplus neutrons, and moderators such as graphite or heavy water, which slow them down, it is the basis of the nuclear reactor.

  34. Nuclear fusion is the combining of two light nuclei to form a heavy nucleus, for example \[^{2}*{1}\\mathrm{H} + ^{3}*{1}\\mathrm{H} \\longrightarrow\\ ^{4}*{2}\\mathrm{He} + ^{1}*{0}n + 17.6\\ \\text{MeV}\] It needs a temperature of about \(10^{7}\\ \\text{K}\) and high pressure to overcome the electrostatic repulsion between the nuclei, which is why it is called a thermo-nuclear reaction.

  35. The sun and the stars are powered by fusion, four hydrogen nuclei forming a helium nucleus with the release of \(26.7\\ \\text{MeV}\). For the same mass of fuel, fusion releases much more energy than fission, since a given mass contains far more light nuclei than heavy ones.

  36. In both fission and fusion, the energy released comes from the same source — the loss in mass — through \(E = (\\Delta m)c^{2}\), as FC 12.8 shows.

Flowchart

FC 12.8 — The Spine Of The Chapter, From Mass Loss To Energy Unstable or bombarded nucleus → nuclear change takes place: radioactive decay, nuclear fission or nuclear fusion → total mass of the products is less than the total mass of the reactants → this difference is the loss in mass Δm → energy released E = (Δm)c² → energy appears as kinetic energy of the products, γ-rays, heat and light → used constructively in a reactor or destructively in a bomb

12.24 Key Terms

  1. Alpha particle — a helium nucleus, made of two protons and two neutrons, emitted by a radioactive nucleus; charge \(+3.2 \\times 10^{-19}\\ \\text{C}\), mass \(6.68 \\times 10^{-27}\\ \\text{kg}\).

  2. Atomic mass unit (a.m.u.) — a unit of mass equal to one-twelfth the mass of one atom of carbon-12, that is \(1.66 \\times 10^{-27}\\ \\text{kg}\), equivalent to \(931\\ \\text{MeV}\) of energy.

  3. Atomic number (Z) — the number of protons in the nucleus of an atom, equal to the number of electrons in a neutral atom; a count, with no unit.

  4. Background radiations — the radiations to which everyone is exposed even when no radioactive source is present, coming from inside the body and from the surroundings; dose measured in sievert (Sv).

  5. Beta particle — a fast moving electron created and emitted by a nucleus when a neutron changes into a proton; charge \(-1.6 \\times 10^{-19}\\ \\text{C}\), rest mass \(9.1 \\times 10^{-31}\\ \\text{kg}\).

  6. Carbon dating — finding the age of once-living material from the rate of decay of carbon-14 in it.

  7. Daughter nucleus — the new nucleus left behind after a parent nucleus has emitted a radiation.

  8. Electron — the light negatively charged particle that revolves around the nucleus; charge \(-1.6 \\times 10^{-19}\\ \\text{C}\), mass \(9.1 \\times 10^{-31}\\ \\text{kg}\).

  9. Gamma radiation — an uncharged electromagnetic wave of wavelength about \(10^{-4}\\ \\text{nm}\) emitted by an excited nucleus, travelling at \(3 \\times 10^{8}\\ \\text{m s}^{-1}\).

  10. Isobars — atoms of different elements having the same mass number \(A\) but different atomic number \(Z\).

  11. Isotones — atoms having the same number of neutrons but different atomic number \(Z\) and different mass number \(A\).

  12. Isotopes — atoms of the same element having the same atomic number \(Z\) but different mass number \(A\).

  13. Mass number (A) — the total number of nucleons, that is protons plus neutrons, in the nucleus of an atom; a count, with no unit.

  14. Neutron — the uncharged particle inside the nucleus; charge zero, mass \(1.67 \\times 10^{-27}\\ \\text{kg}\).

  15. Nuclear change — a change in the number of nucleons inside the nucleus, involving energies of the order of a few MeV.

  16. Nuclear energy — the energy released when mass is lost in a nuclear change, given by \(E = (\\Delta m)c^{2}\) and measured in joule (J), or in MeV for a single reaction.

  17. Nuclear fission — the splitting of a heavy nucleus into two lighter nuclei of nearly the same size when it is bombarded with slow neutrons, releasing nearly \(190\\ \\text{MeV}\) per fission.

  18. Nuclear fusion — the combining of two light nuclei to form a heavier nucleus at very high temperature and pressure, releasing a large amount of energy.

  19. Nuclear waste — the radioactive material, such as spent fuel rods, left over after use, which must be stored safely rather than dumped.

  20. Nucleons — the protons and neutrons together, being the particles that make up the nucleus.

  21. Nucleus — the small dense positively charged core at the centre of an atom, of size \(10^{-15}\\ \\text{m}\) to \(10^{-14}\\ \\text{m}\), containing the protons and neutrons.

  22. Parent nucleus — the original nucleus that undergoes a decay and emits a radiation.

  23. Proton — the positively charged particle inside the nucleus; charge \(+1.6 \\times 10^{-19}\\ \\text{C}\), mass \(1.67 \\times 10^{-27}\\ \\text{kg}\).

  24. Radioactive isotopes — unstable isotopes, having many more neutrons than protons, which decay by emitting radiations.

  25. Radioactivity — the spontaneous emission of \(\\alpha\) or \(\\beta\) and \(\\gamma\) radiations from the nucleus of an atom during its decay; a nuclear phenomenon, unaffected by physical or chemical treatment.

  26. Radio isotopes — radioactive isotopes of elements of atomic number less than 82, used in medicine, science and industry.

  27. Stationary orbits — the fixed shells in which electrons revolve around the nucleus without radiating out energy.

  28. Tracers — weak radioactive isotopes introduced into a body or a plant so that their path can be followed from outside.

  29. Transmutation — the conversion of the nucleus of one element into that of another, for example by bombarding it with alpha particles.

12.25 Points To Ponder

  1. The decay of any single nucleus is completely unpredictable — there is no law that says when it will happen. Yet a large sample of the same substance loses its activity in a perfectly regular way. How can something made up entirely of unpredictable events behave so dependably when there is enough of it?

  2. Nothing we can do to a substance — heating it, freezing it, dissolving it, putting it in a strong magnetic field — changes its rate of decay by the smallest amount. What does this stubbornness suggest about how strongly a nucleus is held together, compared with how strongly an atom holds on to its outer electrons?

  3. A nucleus becomes unstable when it holds far more neutrons than protons. But neutrons carry no charge and do not repel anything. Why should having too many of an uncharged particle make a nucleus unstable, when it is the protons that push each other apart?

  4. Suppose you had to design a container to hold a source giving out all three radiations, using paper, aluminium and lead. Would it make any difference which material you placed on the inside and which on the outside? Think about what each material stops, and about what a \(\\beta\) particle produces when it is stopped by a heavy metal.

  5. For the same mass of fuel, fusion releases far more energy than fission, needs a fuel available in abundance from sea water, and leaves behind no radioactive waste. Fission runs power stations today; fusion does not. What does this gap tell you about the difference between understanding a process and being able to use it?

  6. Carbon dating rests on the assumption that a living plant takes in carbon-14 at a steady rate from its surroundings, and stops at death. What would happen to the ages worked out by this method if the amount of carbon-14 in the atmosphere had not stayed steady over the past few thousand years?

12.26 Objective Questions

  1. (MCQ) The number of neutrons in the nucleus \(^{37}\_{17}\\mathrm{Cl}\) is : (1 mark)
    1. 17
    2. 20
    3. 37
    4. 54
  2. (MCQ) Which of the following pairs are isotones ? (1 mark)
    1. \(^{23}*{11}\\mathrm{Na}\) and \(^{23}*{12}\\mathrm{Mg}\)
    2. \(^{14}*{6}\\mathrm{C}\) and \(^{14}*{7}\\mathrm{N}\)
    3. \(^{23}*{11}\\mathrm{Na}\) and \(^{24}*{12}\\mathrm{Mg}\)
    4. \(^{12}*{6}\\mathrm{C}\) and \(^{13}*{6}\\mathrm{C}\)
  3. (MCQ) When a nucleus emits a \(\\beta\)-particle, its atomic number : (1 mark)
    1. decreases by 1
    2. increases by 1
    3. decreases by 2
    4. remains unchanged
  4. (MCQ) The radiation which is not deflected by a magnetic field is : (1 mark)
    1. \(\\alpha\)-particles
    2. \(\\beta\)-particles
    3. \(\\gamma\)-radiations
    4. cathode rays
  5. (MCQ) One atomic mass unit is equivalent to an energy of : (1 mark)
    1. 931 eV
    2. 931 keV
    3. 931 MeV
    4. 931 J
  6. (MCQ) In a nuclear reactor, the function of a moderator is to : (1 mark)
    1. absorb the surplus neutrons
    2. slow down the neutrons
    3. supply fresh neutrons
    4. cool the reactor core
  7. (MCQ) The energy released in the fission of one nucleus of \(^{235}\_{92}\\mathrm{U}\) is nearly : (1 mark)
    1. 1.9 MeV
    2. 19 MeV
    3. 190 MeV
    4. 1900 MeV
  8. (MCQ) Of the three radiations, the one with the greatest specific charge is : (1 mark)
    1. the \(\\alpha\)-particle
    2. the \(\\beta\)-particle
    3. the \(\\gamma\)-radiation
    4. all three have the same specific charge
  9. (Assertion–Reason) Assertion (A) : The rate of decay of a radioactive substance does not change when the substance is strongly heated. Reason (R) : Radioactivity is a property of the nucleus, while heating affects only the orbital electrons. (1 mark)
    1. Both A and R are true, and R is the correct explanation of A
    2. Both A and R are true, but R is not the correct explanation of A
    3. A is false but R is true
    4. A is true but R is false
  10. (Assertion–Reason) Assertion (A) : A \(\\beta\)-particle is deflected more than an \(\\alpha\)-particle in the same magnetic field. Reason (R) : A \(\\beta\)-particle carries a larger charge than an \(\\alpha\)-particle. (1 mark)
    1. Both A and R are true, and R is the correct explanation of A
    2. Both A and R are true, but R is not the correct explanation of A
    3. A is false but R is true
    4. A is true but R is false
  11. (Match The Following) Match each item in Column A with the correct item in Column B. (4 marks)
Column A Column B
a. \(\\alpha\)-particle i. Fast moving electron created inside the nucleus
b. \(\\beta\)-particle ii. Uncharged particle of mass nearly \(1.67 \\times 10^{-27}\\ \\text{kg}\)
c. \(\\gamma\)-radiation iii. Helium nucleus of charge \(+3.2 \\times 10^{-19}\\ \\text{C}\)
d. Neutron iv. Electromagnetic wave of wavelength about \(10^{-4}\\ \\text{nm}\)
  1. (Fill In The Blanks) Fill in each blank with the correct word, symbol or value. (4 marks)

    1. The number of neutrons in a nucleus is equal to __________.
    2. Atoms of different elements having the same mass number are called __________.
    3. The emission of a __________ radiation changes neither the atomic number nor the mass number of a nucleus.
    4. In a nuclear reactor the surplus neutrons are absorbed by __________ rods.
  2. (True Or False) State whether each statement is true or false. (4 marks)

    1. An electron is present inside the nucleus before a \(\\beta\)-particle is emitted.
    2. Gamma radiations have the maximum ionising power of the three radiations.
    3. The daughter nucleus formed after a \(\\beta\)-emission is an isobar of the parent nucleus.
    4. Nuclear fusion takes place at ordinary temperature and pressure.
  3. (Name The Following) Name the following. (4 marks)

    1. The radiation that is stopped by a sheet of thick paper.
    2. The scientist who discovered radioactivity in 1896.
    3. The radio isotope used in the treatment of cancer.
    4. The gas present in rocks and soil that contributes to background radiation.
  4. (Give One Word Or Phrase) Give one word or phrase for each of the following. (3 marks)

    1. The process in which a heavy nucleus splits into two nearly equal lighter nuclei on being struck by a slow neutron.
    2. The state of matter in which a substance exists at about \(10^{7}\\ \\text{K}\), consisting of free electrons and bare nuclei.
    3. A weak radioactive isotope introduced into the body so that its path can be followed from outside.
  5. (State The SI Unit) State the SI unit of each of the following. (3 marks)

    1. Charge on a \(\\beta\)-particle
    2. Energy released in a nuclear change
    3. Dose of radiation received by a person
  6. (Odd One Out) Pick the odd one out and give a reason for your choice : \(^{12}*{6}\\mathrm{C}\), \(^{13}*{6}\\mathrm{C}\), \(^{14}*{7}\\mathrm{N}\), \(^{14}*{6}\\mathrm{C}\). (2 marks)

  7. (Odd One Out) Pick the odd one out and give a reason for your choice : cadmium rods, graphite, heavy water, lead apron. (2 marks)

12.27 Check Your Concepts

  1. (Short Answer) Define the term mass number, and state the relation connecting it with the atomic number and the number of neutrons. (2 marks)

  2. (Short Answer) State what is meant by isotopes, and give one example of a pair of isotopes with their symbols. (2 marks)

  3. (Short Answer) Give a reason why an \(\\alpha\)-particle has a much smaller specific charge than a \(\\beta\)-particle. (2 marks)

  4. (Short Answer) State Einstein’s mass–energy equivalence relation, naming each symbol in it together with its unit. (2 marks)

  5. (Short Answer) Why is a \(\\gamma\)-radiation not deflected by an electric field, although an \(\\alpha\)-particle and a \(\\beta\)-particle both are ? (2 marks)

  6. (Short Answer) Differentiate between isobars and isotones, stating clearly the basis of the difference. (3 marks)

  7. (Short Answer) Explain why an \(\\alpha\)-particle has the maximum ionising power but the least penetrating power of the three radiations. (3 marks)

  8. (Short Answer) ‘Radioactivity is a nuclear phenomenon.’ State the observation that led to this conclusion, and explain how it rules out the orbital electrons. (3 marks)

  9. (Short Answer) State three safety measures that a person working with radioactive materials must follow, and give a reason for each. (3 marks)

  10. (Short Answer) What changes take place in the nucleus of an atom when it emits (a) an \(\\alpha\)-particle, and (b) a \(\\beta\)-particle ? Give one example of each, written as a nuclear equation. (4 marks)

  11. (Short Answer) Describe the observation made by Henry Becquerel, and state the conclusion he drew from it. (4 marks)

  12. (Short Answer) What is meant by a chain reaction ? Explain how a chain reaction is kept under control in a nuclear reactor, naming the materials used. (4 marks)

  13. (Long Answer) State the three types of harmful biological effects of nuclear radiations, giving one example of each, and state which of them is not limited to the person exposed. (5 marks)

  14. (Long Answer) What are radio isotopes ? Describe one use of radio isotopes in each of the medical, scientific and industrial fields, naming the isotope or the radiation used in each case. (5 marks)

12.28 Diagrams, Graphs & Constructions

  1. (Diagram) Draw a labelled diagram of the model of the sodium atom \(^{23}\_{11}\\mathrm{Na}\), showing the contents of the nucleus and the distribution of electrons in the shells, as in Fig. 12.2-3. (4 marks)

  2. (Diagram) Fig. 12.7-2 shows a radioactive source in a thick lead container placed between two charged plates. Redraw the arrangement and complete it by showing and labelling the paths of the \(\\alpha\), \(\\beta\) and \(\\gamma\) radiations. State which path shows the largest deflection. (4 marks)

  3. (Diagram) Fig. 12.7-1 shows the same beam passing through a magnetic field acting into the plane of the paper. Draw the three paths, label them, and explain how you decided the direction in which each one bends. (4 marks)

  4. (Diagram) Draw a labelled diagram to show the fission of a \(^{235}\_{92}\\mathrm{U}\) nucleus by a slow neutron, on the lines of Fig. 12.19-1, labelling the two fragments and the neutrons released. (4 marks)

  5. (Diagram) Sketch a branching diagram of a chain reaction for three stages, starting from a single neutron and taking three neutrons to be released in each fission. State how many fissions occur in the third stage. (3 marks)

  6. (Diagram) Fig. 12.16-1 shows a radioactive source stored in a thick lead container with a narrow opening. Identify the labelled parts, and explain why the radiations come out only through the opening. (3 marks)

12.29 Practice With Numericals

  1. (Numerical) A nucleus is written as \(^{207}\_{82}\\mathrm{Pb}\). Find the number of protons, the number of neutrons and the number of electrons in its neutral atom. (2 marks)

  2. (Numerical) The charge on a \(\\beta\)-particle is \(1.6 \\times 10^{-19}\\ \\text{C}\) and its rest mass is \(9.1 \\times 10^{-31}\\ \\text{kg}\). Calculate its specific charge. (2 marks)

  3. (Numerical) In a nuclear reaction the loss in mass is \(0.05\\ \\text{a.m.u.}\) Calculate the energy released, in MeV. Take \(1\\ \\text{a.m.u.} = 931\\ \\text{MeV}\). (2 marks)

  4. (Numerical) Calculate the energy released, in joule, when a mass of \(3\\ \\text{g}\) is lost in a nuclear change. Express your answer also in kWh. Take \(c = 3 \\times 10^{8}\\ \\text{m s}^{-1}\) and \(1\\ \\text{kWh} = 3.6 \\times 10^{6}\\ \\text{J}\). (3 marks)

  5. (Numerical) In the fission reaction \[^{235}*{92}\\mathrm{U} + ^{1}*{0}n \\longrightarrow\\ ^{90}*{38}\\mathrm{Sr} + ^{143}*{54}\\mathrm{Xe} + x,^{1}\_{0}n + \\text{energy}\] find the value of \(x\), showing how the mass number and the atomic number balance. If the loss in mass in this reaction is \(0.22\\ \\text{a.m.u.}\), calculate the energy released in MeV. (4 marks)

  6. (Numerical) A nucleus \(^{232}*{90}\\mathrm{Th}\) decays through a series of emissions and finally becomes \(^{208}*{82}\\mathrm{Pb}\). Calculate the number of \(\\alpha\)-particles and the number of \(\\beta\)-particles emitted in the whole series. (4 marks)

  7. (Numerical) In the core of a reactor, \(1 \\times 10^{18}\) fissions take place every second, and each fission releases \(3 \\times 10^{-11}\\ \\text{J}\) of energy. Calculate the energy released per second, and hence state the power of the reactor in watt. (4 marks)