Chapter 13: Nuclear Chemistry and Radioactivity
In the mid-1890s, a Paris laboratory held a puzzle that no one had asked for. Antoine Henri Becquerel was studying uranium salts, hoping to show that they would glow and give out rays only after being charged up by sunlight. His plan was simple: leave the salt in the sun, then place it on a photographic plate wrapped in thick black paper, and see whether anything reached the plate through the wrapping.
Then the Paris sky clouded over. With no sunlight for days, Becquerel put the wrapped plates and the uranium salt away together in a dark drawer. When he finally developed those plates, he expected almost nothing. Instead, the outline of the salt was printed on them, strongly and clearly.
The rays had not come from sunlight. They had not come from any chemical reaction he had set up. They were coming out of the uranium itself, steadily, whether the sample was warm or cold, lit or in the dark. Something inside the atom was changing on its own.
That drawer full of fogged plates opened an entirely new branch of chemistry — one concerned not with electrons and bonds, but with the tiny, dense nucleus at the centre of every atom, and with the reactions in which one element quietly turns into another. The core idea is sketched in Fig. 13.0.
Diagram
Fig. 13.0 – Split illustration: (left) wrapped photographic plate in a dark drawer with a dish of uranium salt on it, showing invisible rays passing through the paper and darkening the plate; (right) a magnified atom with a tiny dense nucleus of protons and neutrons at the centre, labelled “nucleus (~10⁻¹⁵ m)” and an outer electron region labelled “~10⁻¹⁰ m”, with an arrow marked “change happens here”.
As you work through this chapter, keep these five questions in mind — each one will be answered as the ideas behind them are built up, section by section.
13.1 Introduction: Nuclear Chemistry is a Branch of Physical Chemistry
Almost every reaction you have studied so far involves electrons. Bonds break, bonds form, electrons shift from one atom to another — but through all of it, the nucleus of each atom stays exactly as it was. Sodium remains sodium; oxygen remains oxygen.
Nuclear chemistry begins where that rule ends. It is the branch of physical chemistry that studies reactions in which the atomic nucleus itself changes. In such a reaction, the number of protons or neutrons in a nucleus is altered, so one element can genuinely be converted into another.
This branch began with the discovery of natural radioactivity by the physicist Antoine Henri Becquerel (1852–1908). Once it was clear that a nucleus could change on its own, a natural question followed: what holds a nucleus together in the first place? Much of the research of the twentieth century was directed towards understanding these forces that bind the nucleus.
Why Nuclear Chemistry Matters
Nuclear reactions are not a laboratory curiosity; they run quietly through several sciences at once. Geologists explore nuclear reactions to understand the changes that have occurred on the earth, because the slow decay of unstable nuclei in rocks acts as a clock built into the rock itself.
Astronomers study the nuclear reactions taking place in stars. The light and heat of the Sun are not produced by burning in the chemical sense at all — they come from nuclei combining.
In biology and medicine, what matters most is the interaction of the radiation emitted from nuclear reactions with the living system. The same radiation that can damage healthy tissue can, when carefully aimed and measured, destroy a tumour or reveal the working of an organ.
The Reactions We Will Meet
Four kinds of nuclear reaction appear in this chapter: radioactive decay, artificial transmutation, nuclear fission and nuclear fusion. They differ in whether the change happens on its own or is forced on the nucleus from outside, and in whether a nucleus breaks apart or nuclei join together. The overall plan is set out in FC 13.1.
Flowchart
FC 13.1 – Nuclear reactions covered in this chapter: nuclear reactions → branch A: spontaneous (natural) → radioactive decay → α-decay / β-decay / γ-emission; branch B: non-spontaneous (man-made) → artificial transmutation → induced radioactivity → nuclear fission (heavy nucleus splits) and nuclear fusion (light nuclei join)
Since every one of these processes is a change in the nucleus, we first need a clear picture of the nucleus itself. The structure of the atom was studied in Chapter 4, and the next part of this section recalls the parts of that picture we will use constantly.
13.1.1 Similarity Between the Solar System and Structure of Atom
Our solar system is made up of the Sun and the planets. The Sun sits at the centre, and the planets move around it under the force of gravity — one central massive body, and much lighter bodies held in place by an attraction reaching out from it.
An atomic system is arranged in an analogous way, as drawn in Fig. 13.1a. Here the forces which hold the nucleus and the extranuclear electrons together are attractive electrostatic forces: the positive nucleus attracts the negative electrons around it. The geometry is similar, but the force is different — gravity in one case, electrostatic attraction in the other.
Diagram
Fig. 13.1a – Side-by-side analogy: (left) the Sun at the centre with planets in orbits, arrows labelled “gravitational attraction”; (right) a nucleus at the centre with fast-moving electrons in the surrounding region, arrows labelled “electrostatic attraction”. Label the central bodies “Sun” and “nucleus (protons + neutrons)”.
The Two Regions of an Atom
As studied in Chapter 4, the atom consists of a tiny central core called the nucleus, made of protons and neutrons, and a surrounding region of space occupied by fast moving electrons.
These two regions are on utterly different scales. The radius of the nucleus is of the order of \(10^{-15}\ \text{m}\), whereas that of the outer sphere is of the order of \(10^{-10}\ \text{m}\). The size of the outer sphere is therefore \(10^{5}\) times larger than the nucleus, which means there is a large vacant space outside the nucleus — the atom is mostly empty room.
How a Nuclide is Written
The composition of an atom is denoted as \(^{A}_{Z}\mathrm{X}\), where X is the symbol of the element. Here \(Z\) equals the number of protons and is called the atomic number of the element (refer to Chapter 4, section 4.2).
\(A\) is the mass number, the sum of the number of protons and neutrons in an atom, and \(N\) is the number of neutrons in an atom. The three are linked by a relation you will use in almost every problem in this chapter:
\[A = Z + N\]
The number of neutrons can therefore be obtained by subtracting the atomic number \(Z\) from the mass number \(A\):
\[N = A - Z\]
Reading the symbol is a matter of position, shown in Fig. 13.1b. The atomic number, appearing as a subscript to the left of the elemental symbol, gives the number of protons in the nucleus. The mass number, written as a superscript to the left of the element symbol, gives the total number of nucleons — that is, the sum of protons (p) and neutrons (n).
The most common isotope of carbon, for example, has 12 nucleons: 6 protons and 6 neutrons. Written out, it is \(^{12}_{\ 6}\mathrm{C}\), or carbon-12.
Diagram
Fig. 13.1b – The symbol ¹²₆C drawn large, with a labelled arrow from the superscript reading “mass number (A) = 12 nucleons” and a labelled arrow from the subscript reading “atomic number (Z) = 6 protons”; alongside, a small tally: 6 protons + 6 neutrons = 12 nucleons; caption “Carbon-12”.
As you know, atoms with identical atomic numbers but different mass numbers are called isotopes. The nucleus of a specific isotope is called a nuclide. The distinction is worth holding on to: when we say “isotope” we are usually thinking of the whole atom, but when we say “nuclide” we mean the nucleus alone, with its own definite count of protons and neutrons.
Where the Mass Sits
The charge on the nucleus is \(+Ze\) and that of the outer sphere is \(-Ze\), where \(e\) is the magnitude of the electronic charge. The two are equal and opposite, so the atom as a whole is electrically neutral.
The mass of an electron is negligible — about \(1/1837\)th of the mass of a proton — in comparison to a proton or a neutron. The entire mass of the atom is therefore concentrated in its nucleus, even though the nucleus occupies almost none of the atom’s volume.
Squeezing nearly all the mass into nearly none of the volume has a striking consequence. The density of the nucleus is considerably higher than that of anything we handle — typically \(10^{5}\) times the density of ordinary matter.
The Size of a Nucleus
The radius of the nucleus is given by
\[R = R_0 A^{1/3}\]
where \(R\) is the nuclear radius, \(A\) is the mass number, and \(R_0\) is a constant common to all nuclei whose value is \(1.33 \times 10^{-15}\ \text{m}\). Notice what this expression says: the radius depends on nothing except the number of nucleons packed in.
The volume of the nucleus is proportional to the cube of its radius, \(V \propto R^{3}\), and since \(R \propto A^{1/3}\), cubing gives \(R^3 \propto A\). Hence
\[V \propto A\]
This picture — a small, dense, positively charged core described completely by \(Z\) and \(A\) — is the starting point for everything that follows. The next section sorts nuclides into families on the basis of how many protons and neutrons they contain, and this classification will be the language used through the rest of the chapter.
13.2 Classification of Nuclides
Once we describe a nucleus by just two numbers — its proton count \(Z\) and its neutron count \(N\) — a natural way of sorting nuclei appears. Two nuclei may share their proton count, or their neutron count, or their total nucleon count, and each kind of sharing gives a family with its own name and its own uses.
Nuclides can be classified in two quite different ways: by the numbers of nucleons they contain, and by whether they are stable or radioactive. Both schemes are set out in FC 13.2, and both are used constantly for the rest of this chapter.
Flowchart
FC 13.2 – Classification of nuclides: nuclides → branch A: on the basis of number of nucleons → isotopes (same Z) / isobars (same A) / mirror nuclei (special isobars, Z and N interchanged, differing by 1) / isotones (same N) / nuclear isomers (same Z and N, different energy state); branch B: on the basis of nuclear stability → stable nuclides / unstable (radioactive) nuclides → spontaneous change → new nuclides + radiation
13.2.1 Classification on the Basis of Number of Nucleons
This classification is based on the number of neutrons and protons constituting the nucleus. Remember that the word nuclide refers to the atomic nucleus alone, without any relation to the outer sphere of electrons — so all these families are decided by the nucleus only.
On this basis, nuclides are classified as follows.
Isotopes — Same Number of Protons
Isotopes are nuclides which have the same number of protons but a different number of neutrons. Because the proton count fixes the element, all isotopes of a set belong to the same element; only the mass number differs.
For example, \(^{22}_{11}\mathrm{Na}\), \(^{23}_{11}\mathrm{Na}\) and \(^{24}_{11}\mathrm{Na}\) are isotopes, the number of neutrons in each being 11, 12 and 13 respectively. Each has 11 protons, since each is sodium; subtracting 11 from 22, 23 and 24 gives those neutron numbers.
Isobars — Same Mass Number
Isobars are nuclides which have the same mass number but different numbers of protons and neutrons. Here the total nucleon count matches, but the split between protons and neutrons is different — so isobars are always different elements.
For example, in \(^{14}_{6}\mathrm{C}\) and \(^{14}_{7}\mathrm{N}\) the number of neutrons is 8 and 7 respectively, while both have \(A = 14\). Likewise \(^{3}_{1}\mathrm{H}\) and \(^{3}_{2}\mathrm{He}\) are a pair of isobars. Several such pairs are collected in Table 13.1.
Table 13.1 : Isobars
| Isobars of | Nuclides |
|---|---|
| \(A = 3\) | \(^{3}_{1}\mathrm{H}\) (\(N = 2\)), \(^{3}_{2}\mathrm{He}\) (\(N = 1\)) |
| \(A = 14\) | \(^{14}_{6}\mathrm{C}\) (\(N = 8\)), \(^{14}_{7}\mathrm{N}\) (\(N = 7\)) |
| \(A = 24\) | \(^{24}_{11}\mathrm{Na}\) (\(N = 13\)), \(^{24}_{12}\mathrm{Mg}\) (\(N = 12\)) |
Read down any row of Table 13.1 and you will see the same total, \(A\), split two ways. In the \(A = 3\) row, one nucleus carries 1 proton and 2 neutrons; the other carries 2 protons and 1 neutron. This is worth looking at closely, because that particular pair also belongs to the next family.
Diagram
Fig. 13.2a – Three panels of nucleon “grids” (filled circles for protons, open circles for neutrons): panel 1 labelled “isotopes” showing ²²Na, ²³Na, ²⁴Na with the proton block identical and the neutron block growing; panel 2 labelled “isobars” showing ¹⁴C (6p, 8n) and ¹⁴N (7p, 7n) with equal total circles; panel 3 labelled “isotones” showing ¹³C (6p, 7n) and ¹⁴N (7p, 7n) with the neutron block identical.
Mirror Nuclei
Mirror nuclei are isobars in which the numbers of protons and neutrons differ by 1 and are interchanged. In other words, whatever one nucleus has as protons, the other has as neutrons, and the other way round.
Examples are \(^{3}_{1}\mathrm{H}\) and \(^{3}_{2}\mathrm{He}\), and \(^{13}_{6}\mathrm{C}\) and \(^{13}_{7}\mathrm{N}\). In the first pair, 1 proton with 2 neutrons faces 2 protons with 1 neutron; in the second, 6 protons with 7 neutrons faces 7 protons with 6 neutrons. Every mirror pair is therefore also a pair of isobars, but only a few isobars qualify as mirror nuclei.
Isotones — Same Number of Neutrons
Isotones are nuclides having the same number of neutrons but a different number of protons, and hence different mass numbers. Since the neutron count is what matches, isotones are also different elements.
The elements carbon, nitrogen, sodium and magnesium provide two standard examples: \(^{13}_{6}\mathrm{C}\) and \(^{14}_{7}\mathrm{N}\), and \(^{23}_{11}\mathrm{Na}\) and \(^{24}_{12}\mathrm{Mg}\). Check the first pair: \(13 - 6 = 7\) and \(14 - 7 = 7\), so both have 7 neutrons. In the second pair, \(23 - 11 = 12\) and \(24 - 12 = 12\), so both have 12 neutrons.
Nuclear Isomers
The nuclides with the same number of protons (\(Z\)) and neutrons (\(N\)), or the same mass number (\(A\)), which differ in energy states are called nuclear isomers. This family is unusual: the two nuclides are made of exactly the same nucleons in exactly the same numbers, so they are the same nuclide by composition — yet they are not identical, because one holds more energy than the other.
For example, \(^{60\mathrm{m}}\mathrm{Co}\) and \(^{60}\mathrm{Co}\) are nuclear isomers. An isomer of higher energy is said to be in the metastable state, and it is indicated by writing m after the mass number. The energy difference between the two states is shown in Fig. 13.2b.
Diagram
Fig. 13.2b – Nuclear energy-level sketch: an upper horizontal line labelled “metastable state, ⁶⁰ᵐCo (higher energy)” and a lower line labelled “ground state, ⁶⁰Co”, with a downward arrow between them labelled “excess energy given out as radiation”.
13.2.2 Classification on the Basis of Nuclear Stability
The second scheme asks a different question: will this nucleus stay as it is, or will it change on its own? On this basis nuclides are of two kinds.
Stable Nuclides
In a stable nuclide, the number of electrons and the location of electrons may change in the outer sphere, but the number of protons and neutrons in the nucleus is unchanged. This is exactly what happens in ordinary chemistry: when sodium metal reacts and becomes \(\mathrm{Na^{+}}\), an electron has left the outer region, but the sodium nucleus is untouched. Ionisation, bonding and every reaction you have studied so far leave stable nuclei exactly as they were.
Unstable or Radioactive Nuclides
Unstable or radioactive nuclides are those which undergo a spontaneous change in their composition, with emission of radiation, forming new nuclides. “Spontaneous” is the key word: nothing has to be done to the sample. The nucleus itself is not a workable arrangement of nucleons, and it rearranges on its own, throwing out a particle or radiation and leaving behind a different nuclide — often a different element.
This division immediately raises the question the next section takes up: what makes one nucleus a workable arrangement of nucleons and another one not? In other words, which combinations of \(Z\) and \(N\) are stable, and why?
13.3 Nuclear Stability
Why are some nuclei stable while others are radioactive and undergo spontaneous change? To answer this question we need to know the factors governing the stability of the nucleus.
Four such factors are taken up in turn: whether the proton and neutron counts are even or odd, the ratio of neutrons to protons, certain special “magic” counts of nucleons, and the attractive force that acts between nucleons. A fifth idea, nuclear binding energy, then puts a number on stability so that different nuclides can be compared directly.
13.3.1 Even-Odd Nature of Proton Number (\(Z\)) and Neutrons (\(N\))
Nature has already carried out a survey for us. If we count how many stable, naturally occurring nuclides fall into each even/odd combination of \(Z\) and \(N\), a strong pattern appears (Table 13.2).
Table 13.2 : Distribution of naturally occurring nuclides.
| Number of protons \(Z\) | Number of neutrons \(N\) | Number of such nuclides |
|---|---|---|
| Even | Even | 165 |
| Even | Odd | 55 |
| Odd | Even | 50 |
| Odd | Odd | 04 |
The distribution of naturally occurring nuclides in accordance with even/odd values of \(N\) and \(Z\) is useful to understand the stability of nuclides. Three inferences follow from the four rows.
a. The nuclides with even \(Z\) and even \(N\) constitute 85% of the earth’s crust. This indicates that nuclides with even \(Z\) and even \(N\) are stable. They tend to form proton–proton and neutron–neutron pairs, and this pairing imparts stability to the nucleus.
b. The number of stable nuclides with either \(Z\) or \(N\) odd is about one third of those in which both are even. This suggests that these nuclides are less stable than those having even numbers of protons and neutrons, because in them one nucleon has no partner. Further, the counts for nuclides with odd \(Z\) (50) and odd \(N\) (55) are nearly the same, which indicates that protons and neutrons behave similarly in respect of stability.
c. The stable naturally occurring nuclides with odd \(Z\) and odd \(N\) are only four. Such a small number is itself an indication of instability, and it can be attributed to the presence of two unpaired nucleons — one unpaired proton and one unpaired neutron. This also tells us that the pairing of neutrons and of protons happens separately: nucleon pairing does not take place between a proton and a neutron. Only 2% of the earth’s crust consists of such nuclides.
13.3.2 Neutron to Proton Ratio (\(N/Z\))
i. Figure 13.1 shows a plot of the neutron number (\(N\)) as a function of the proton number (\(Z\)) in stable nuclides. A large number of elements have several stable isotopes, so instead of a single thin line, the points spread out and the curve appears as a belt or zone, called the stability zone. All the stable nuclides fall within this zone, and the nuclides which form outside such a belt are radioactive. The straight line lying below the belt represents the ratio \(N/Z\) equal to unity.
Diagram
Fig. 13.1 – Plot of neutron number N (y-axis, 0–140) against proton number Z (x-axis, 0–100); shade the belt of stable nuclei, mark separate symbols for stable nuclei, naturally occurring radioisotopes and artificially produced radioisotopes; draw the straight line N = Z below the belt and label it; add diagonal constant-A lines labelled 10, 30, 50 … 230; label the belt “stability zone”.
ii. As may be noticed for nuclei lighter than \(^{40}_{20}\mathrm{Ca}\), the straight line \(N = Z\) passes through the belt itself. The lighter nuclides are therefore stable when their \(N/Z\) ratio is nearly 1 — that is, when protons and neutrons are present in roughly equal numbers.
iii. The \(N/Z\) ratio for stable nuclides heavier than calcium gives a curved appearance to the belt, with a gradual increase of \(N/Z\) to values greater than 1. The heavier nuclides therefore need a larger number of neutrons than protons to attain stability.
What is the reason for this? The heavier nuclides, with their increasing number of protons, suffer large coulombic repulsions, since every proton pushes every other proton away. With an increased number of neutrons the protons within the nucleus get more separated from one another, and this greater separation weakens the repulsion and renders the nucleus stable. Neutrons, carrying no charge, add attraction without adding any repulsion.
13.3.3 Magic Numbers
Some nucleon counts are unusually favourable. The nuclei with 2, 8, 20, 28, 50, 82 and 126 neutrons or protons are particularly stable and abundant in nature, and these counts are called magic numbers.
A nuclide may be magic in one count or in both. Lead, \(^{208}_{82}\mathrm{Pb}\), has two magic numbers: 82 protons and 126 neutrons. Note how this fits with the earlier ideas — 82 and 126 are both even, so the pairing argument of section 13.3.1 works in the same direction.
13.3.4 Nuclear Potential
Within the nucleus, all protons are positively charged. What are the consequences of the Coulomb repulsions between these protons?
The distance between two protons present in the nucleus is typically \(10^{-15}\ \text{m}\). Coulomb repulsion grows rapidly as charges are brought closer together, so at such a tiny separation the repulsion between two protons has to be large. If the coulomb repulsion had been the only force in operation, the nucleus would not exist at all — it would blow itself apart the moment it formed.
However, the nucleus does exist. This simple observation is the evidence for something more: there must be nuclear forces of attraction between all the nucleons, holding them together within the nucleus. These are attractions between proton and proton, between neutron and neutron, and between proton and neutron.
These attractive forces are independent of the charge on the nucleons. Hence the p–p, n–n and p–n attractions are all equal — the neutral neutron is attracted just as strongly as the charged proton, which shows at once that this is not an electrical force. These attractive forces operate only over a short range within the nucleus.
The p–p, n–n and p–n attractions constitute what is called the nuclear potential, which is responsible for nuclear stability.
The short range of the nuclear attraction is the quiet reason behind the whole of section 13.3.2. Attraction reaches only to a nucleon’s near neighbours, while repulsion between protons reaches right across the nucleus. As a nucleus grows larger, repulsion therefore accumulates faster than attraction, and extra neutrons are needed to keep the arrangement workable.
13.3.5 Nuclear Binding Energy and Mass Defect
The binding energy of a nucleus gives a measure of how strongly the nucleons are held together within the nucleus. It converts a qualitative statement — “this nucleus is tightly bound” — into a number in energy units that can be compared from one nuclide to another.
The energy required for holding the nucleons together within the nucleus of an atom is called the nuclear binding energy. It is defined as the energy required to break the nucleus into its constituents. Here the binding of electrons to the nucleus has been neglected, since it is smaller by an enormous factor and would not affect the answer.
Mass Defect
Atoms are composed of protons, neutrons and electrons. The actual mass of an atom is observed to be less than the sum of the masses of its constituents. This is a genuinely surprising result: if you weigh the parts and then weigh the assembled whole, the whole comes out lighter.
Protons and neutrons are held in the nucleus together, and during the formation of the nucleus a certain mass is lost. This lost mass is known as the mass defect, \(\Delta m\). An energy equivalent to the mass lost is released during the formation of the nucleus, and this is called the nuclear binding energy.
In general, the exact mass of a nucleus is slightly less than the sum of the exact masses of the constituent nucleons. The difference is called the mass defect, \(\Delta m\).
The conversion of mass into energy is established through Einstein’s equation,
\[E = mc^2\]
where \(m\) is the mass of matter converted into energy \(E\), and \(c\) is the velocity of light. The missing mass has not vanished; it has appeared as the energy given out when the nucleus formed. To pull the nucleus apart again, that same energy must be supplied — which is why one number, the binding energy, answers both descriptions.
Units of Nuclear Masses and Energy
Nuclear mass is usually expressed in the unified mass unit (u), which is exactly \(1/12\)th of the mass of the \(^{12}\mathrm{C}\) atom. Thus
\[1\ \text{u} = \tfrac{1}{12}\ \text{of the mass of a C-12 atom} = 1.66\times10^{-27}\ \text{kg}\]
The energy released (in joules) in the conversion of 1 u of mass into energy is given by the expression
\[E = mc^2 = (1.66\times10^{-27}\ \text{kg})\times(3\times10^{8}\ \text{m s}^{-1})^2\]
Working this out gives an energy of about \(1.49\times10^{-10}\ \text{J}\). Such a number is awkwardly small in joules, so nuclear energy is usually expressed in million electron volt (MeV), where
\[1\ \text{MeV} = 1.602\times10^{-13}\ \text{J}\]
Dividing the energy above by this conversion factor turns joules into MeV, and gives the figure used throughout nuclear chemistry: when mass equal to 1 u is converted into energy, it is 931.4 MeV.
Expression for Nuclear Binding Energy
Consider a nuclide \(^{A}_{Z}\mathrm{X}\) that contains \(Z\) protons and \((A - Z)\) neutrons. Suppose the observed mass of the nuclide is \(m\). The mass of a proton is \(m_p\) and that of a neutron is \(m_n\).
The calculated mass is the total mass of all the separate particles. Since tables usually list masses of neutral atoms rather than of bare nuclei, the \(Z\) electrons are included as well:
\[\text{Calculated mass} = (A-Z)\,m_n + Z\,m_p + Z\,m_e \qquad \dots (13.1)\]
Subtracting the observed mass gives the mass defect:
\[\Delta m = [(A-Z)\,m_n + Z\,m_p + Z\,m_e] - m\]
The next step is only a regrouping. The \(Z\) proton masses and the \(Z\) electron masses are collected together:
\[\Delta m = [(A-Z)\,m_n + Z\,(m_p + m_e)] - m\]
\[\Delta m = [(A-Z)\,m_n + Z\,m_H] - m \qquad \dots (13.2)\]
where \((m_p + m_e) = m_H\) = mass of the H atom. This regrouping is a real convenience: a hydrogen atom is one proton plus one electron, so its tabulated mass already contains both, and we never need the proton and electron masses separately.
Written in the order in which it is usually applied,
\[\Delta m = [Z\,m_H + (A-Z)\,m_n] - m\]
where \(Z\) is the atomic number, \(A\) is the mass number, \((A-Z)\) is the neutron number, \(m_H\) and \(m_n\) are the masses of the hydrogen atom and the neutron respectively, and \(m\) is the mass of the nuclide.
The mass defect \(\Delta m\) is related to the binding energy of the nucleus by Einstein’s equation
\[\Delta E = \Delta m \times c^2\]
where \(\Delta E\) is the binding energy and \(\Delta m\) is the mass defect. Rather than working in kilograms and joules every time, we use the equivalence established above — 1 u corresponds to 931.4 MeV — so that a mass defect in u converts straight into an energy in MeV. The total binding energy is then given by
\[\text{B.E.} = \Delta m\ (\text{u}) \times 931.4\ (\text{MeV per u})\]
\[\text{B.E.} = 931.4\,\{[Z\,m_H + (A-Z)\,m_n] - m\} \qquad \dots (13.3)\]
Binding Energy per Nucleon
The total binding energy of a nucleus containing \(A\) number of nucleons is denoted as B.E. Comparing total binding energies of different nuclides is not very useful on its own, because a big nucleus has more nucleons to bind and will naturally show a larger total. To compare fairly we divide by the number of nucleons. The binding energy per nucleon is then given by
\[\bar{B} = \text{B.E.}\,/\,A \qquad \dots (13.4)\]
Binding Energy per Nucleon and Nuclear Stability
The mean binding energy per nucleon for the most stable isotopes, plotted as a function of mass number, is shown in Fig. 13.2. This single plot is one of the most informative graphs in nuclear chemistry, and it leads to the following inferences.
Diagram
Fig. 13.2 – Graph of binding energy per nucleon (MeV, y-axis 0–10) against number of nucleons A (x-axis 0–250); curve rising steeply through ²H, ³He, ⁴He, mark peaks at ⁴He, ¹²C, ¹⁶O, broad maximum near ⁵⁶Fe (8.79 MeV), then a slow fall through ¹²⁰Sn, ²⁰⁸Pb to ²³⁸U; upward arrows on the left labelled “fusion” and on the right labelled “fission”, both pointing towards the maximum.
(i) Light nuclides (\(A < 30\))
In this region the curve rises steeply, and it carries peaks at \(A\) values that are multiples of 4. For example, \(^{4}_{2}\mathrm{He}\), \(^{12}_{6}\mathrm{C}\) and \(^{16}_{8}\mathrm{O}\) are more stable than their neighbours. Each of these has equal, even numbers of protons and neutrons, so the pairing idea of section 13.3.1 shows up here as visible bumps on a graph.
(ii) Medium mass nuclides (\(30 < A < 90\))
\(\bar{B}\) increases typically from 8 MeV for \(A = 16\) to nearly 8.3 MeV for \(A\) between 28 and 32, and it then remains nearly constant at 8.5 MeV. Beyond this it shows a broad maximum. The nuclides falling on the maximum are the most stable, as they possess high \(\bar{B}\) values. Accordingly \(^{56}\mathrm{Fe}\), with a \(\bar{B}\) value of 8.79 MeV, is the most stable nuclide.
(iii) Heavy nuclides (\(A > 90\))
\(\bar{B}\) decreases from the maximum of 8.79 MeV to 7.7 MeV for \(A \cong 210\). \(^{209}\mathrm{Bi}\) is the heaviest stable nuclide; beyond this, all nuclides are radioactive (\(\alpha\)-emitters).
We now have the factors that decide whether a nucleus can survive as it is. The next question is what an unstable nucleus actually does about it — which is the subject of radioactivity.
13.4 Radioactivity
You know that some elements, such as uranium and radium, are radioactive elements. An element is radioactive if the nuclei of its atoms are unstable — that is, if its nuclides lie outside the stability zone described in the previous section.
An unstable nucleus does not simply sit and wait. The phenomenon in which nuclei spontaneously emit a nuclear particle and gamma radiation, transforming to a different nuclide, is called radioactivity. The elements which undergo such nuclear changes are radioactive elements, and they are often referred to as radioelements.
Two words in that definition deserve attention. Spontaneously means that nothing is done to the sample from outside; the change comes entirely from the nucleus itself. And transforming to a different nuclide means the product is not the starting substance in a new form — its nucleus has a different composition, so in most cases a different element has been produced.
Radioactivity is a phenomenon related to the nucleus. This single statement explains a great deal. It is why radioactivity is not altered by dissolving the substance, by combining it into a compound, or by any of the treatments that change the course of an ordinary chemical reaction, all of which act on the outer electrons only. The general shape of every radioactive event is set out in FC 13.4.
Flowchart
FC 13.4 – The radioactive event: unstable parent nucleus (lies outside the stability belt) → spontaneous nuclear change, no external trigger needed, outer electrons take no part → emission → branch 1: a nuclear particle is thrown out → branch 2: gamma radiation is given out → in each case a different nuclide is left behind (usually a different element), lying closer to the stability belt
The Three Radiations
The radiations emitted by radioactive elements are alpha (\(\alpha\)), beta (\(\beta\)) and gamma (\(\gamma\)) radiations. Earlier you have studied the properties of \(\alpha\), \(\beta\) and \(\gamma\) rays — how they differ in charge, in mass, in penetrating power and in ionising power. Those properties are worth recalling now, because from section 13.6 onwards we shall be tracking exactly what each of them does to the mass number and the atomic number of the parent nucleus.
The next section takes up the question that makes radioactivity measurable: if the change is spontaneous and unpredictable for any single nucleus, how fast does a whole sample decay?
13.5 Radioactive Decay
The probability of decay of a radioelement does not depend on the state of chemical combination, the temperature, the pressure, the presence of a catalyst, or even the age of the nucleus.
Read that sentence again, because it removes at one stroke every handle a chemist normally has on a reaction. A nucleus that has already existed for a thousand years is no more “due” to decay than one formed a second ago. Decay is a matter of probability for each individual nucleus, and that probability is fixed for a given nuclide.
13.5.1 Rate of Decay
The rate of decay of a radioelement denotes the number of nuclei of its atoms which decay in unit time. It is also called the activity of the radioelement.
If \(dN\) is the number of nuclei that decay within a time interval \(dt\), then the rate of decay at any time \(t\) is given by
\[\text{rate of decay (activity)} = -\frac{dN}{dt}\]
Why is the minus sign required? The number of nuclei decreases with time, so \(dN\) is a negative quantity. However, the rate of decay is a positive quantity — we cannot report a negative number of disintegrations. The negative sign is introduced in the rate expression to make the rate positive. The rate of decay is expressed as disintegrations per second (dps).
13.5.2 Rate Law
The rate of decay of a radioelement at any instant is proportional to the number of nuclei (atoms) present at that instant. This is reasonable: if each nucleus has the same fixed chance of decaying in the next second, then twice as many nuclei must give twice as many disintegrations per second. Thus,
\[-\frac{dN}{dt} \propto N, \quad \text{or} \quad -\frac{dN}{dt} = \lambda N \qquad \dots (13.5)\]
Rearranging Eq. (13.5) makes the meaning of \(\lambda\) clear:
\[\therefore \lambda = -\frac{dN}{dt} \times \frac{1}{N}\]
where \(-\dfrac{dN}{dt}\) is the rate of decay at any time \(t\), \(N\) is the number of nuclei present at time \(t\), and \(\lambda\) is the decay constant — that is, the fraction of nuclei decaying in unit time.
Notice that Eq. (13.5) has the same form as a first-order rate law in chemical kinetics, with the number of nuclei \(N\) playing the part of a concentration. Everything that follows in this section is the mathematics of a first-order process, applied to nuclei.
13.5.3 Expression for Decay Constant
Equation (13.5) tells us the rate at an instant. What we usually want instead is how many nuclei are left after a given time, and for that the equation must be integrated.
Rearranging Eq. (13.5) so that the two variables are separated:
\[\frac{dN}{N} = -\lambda\,dt\]
Integrating,
\[\int \frac{dN}{N} = -\int \lambda\,dt\]
The integral of \(dN/N\) is the natural logarithm of \(N\), which is where the logarithm in all decay calculations comes from. On performing the integration, we get
\[\ln N = -\lambda t + C \qquad \dots (13.6)\]
where \(C\) is the constant of integration, whose value is obtained as follows.
Finding the Constant of Integration
Let \(N_0\) be the number of nuclei present at some arbitrary zero time. At time \(t\), the number of nuclei is \(N\). So at \(t = 0\), \(N = N_0\). Substituting in Eq. (13.6), and remembering that \(-\lambda \times 0 = 0\),
\[\ln N_0 = C\]
With this value of \(C\), Eq. (13.6) becomes
\[\ln N = -\lambda t + \ln N_0\]
\[\text{or} \quad \lambda t = \ln N_0 - \ln N = \ln \frac{N_0}{N} \qquad \dots (13.7)\]
The last step uses the logarithm rule \(\ln a - \ln b = \ln (a/b)\). Hence
\[\lambda = \frac{1}{t}\ln \frac{N_0}{N} \qquad \dots (13.8)\]
Changing to Base-10 Logarithms
Log tables and calculators are set up for logarithms to the base 10, so it is convenient to convert. Since \(\ln x = 2.303\log_{10}x\), converting the natural logarithm (ln) to logarithm to the base 10, Eq. (13.8) becomes
\[\lambda = \frac{2.303}{t}\log_{10}\frac{N_0}{N} \qquad \dots (13.9)\]
This is the working equation for most numerical problems on decay. It links four quantities — \(\lambda\), \(t\), \(N_0\) and \(N\) — so that any one of them can be found if the other three are known.
The Exponential Form
Equation (13.7) can also be expressed as \(\ln \dfrac{N}{N_0} = -\lambda t\). Taking the antilogarithm of both sides, we get
\[\frac{N}{N_0} = e^{-\lambda t} \quad \text{or} \quad N = N_0e^{-\lambda t} \qquad \dots (13.10)\]
Equations (13.9) and (13.10) give the decay constant. They contain the same information in two forms: Eq. (13.9) is convenient when \(\lambda\) or \(t\) is to be calculated, and Eq. (13.10) shows directly that the number of nuclei falls away exponentially with time — steeply at first, then more and more slowly.
13.5.4 Half Life of Radioelement (\(t_{1/2}\))
The half life of a radioelement is the time needed for a given number of its nuclei (atoms) to decay exactly to half of its initial value. Each radioisotope has its own half life, denoted by \(t_{1/2}\) and expressed in seconds, minutes, hours, days or years, depending on how fast it decays.
At any time \(t\), the number of nuclei is \(N\). At \(t = 0\), \(N = N_0\). Hence at \(t = t_{1/2}\), by the definition of half life, \(N = N_0/2\). Substitution of these values of \(N\) and \(t\) in Eq. (13.9) gives
\[\lambda = \frac{2.303}{t_{1/2}}\log_{10}\frac{N_0}{N_0/2}\]
The ratio inside the logarithm is simply 2, since \(N_0\) cancels. Therefore
\[\lambda = \frac{2.303}{t_{1/2}}\log_{10}2 = \frac{2.303}{t_{1/2}} \times 0.3010 = \frac{0.693}{t_{1/2}}\]
Hence
\[\lambda = \frac{0.693}{t_{1/2}} \quad \text{or} \quad t_{1/2} = \frac{0.693}{\lambda}\]
The important feature of this result is what is missing from it: \(N_0\) has cancelled out. The half life does not depend on how much of the sample you start with. One gram and one kilogram of the same radioelement take exactly the same time to fall to half. This is what allows \(t_{1/2}\) to be quoted as a fixed property of each nuclide, and it is why \(\lambda\) and \(t_{1/2}\) are two ways of stating the same thing — a large \(\lambda\) means fast decay and therefore a short half life.
13.5.5 Graphical Representation of Decay
i. From Eq. (13.9), taking the logarithmic term to one side, we can write
\[\log_{10}N = \frac{-\lambda}{2.303}\,t + \log_{10}N_0\]
This has the form of a straight line, \(y = mx + c\), with \(\log_{10}N\) as \(y\) and \(t\) as \(x\). A plot of \(\log_{10}N\) versus \(t\) therefore gives a straight line whose slope is \(-\lambda/2.303\), from which \(\lambda\) can be found.
There is a practical difficulty, however: the number of nuclei \(N\) in a sample cannot be counted directly. What an instrument actually measures is the activity. Since \(N \propto \left(-\dfrac{dN}{dt}\right)\), the activity can stand in for \(N\) without changing the shape of the graph. Hence, instead of \(\log_{10}N\) versus \(t\), we plot \(\log_{10}\left(-\dfrac{dN}{dt}\right)\), which is \(\log_{10}(\text{activity})\), versus time. The graph is a straight line as shown in Fig. 13.3.
Diagram
Fig. 13.3 – Straight-line graph with “log activity” on the y-axis and “Time” on the x-axis, sloping downwards from left to right, annotated “slope = −λ/2.303”.
ii. The rate of radioactive decay at any instant is proportional to the number of atoms of the radioactive element present at that instant. It is evident that as decay progresses, the number of radioactive atoms decreases with time, and so does the rate of decay. A plot of the rate, or activity, versus \(t\) is shown in Fig. 13.4.
Diagram
Fig. 13.4 – Decay curve with “Activity” on the y-axis (0–800) and “Time” on the x-axis (0–1500); a smooth curve falling steeply from about 800 at t = 0, flattening as it approaches the time axis without touching it; mark the plotted points.
Fig. 13.4 shows a decrease in the rate of decay, that is, a decrease in the number of radioactive atoms with time. The two figures describe the same process: Fig. 13.4 shows the true exponential fall, while Fig. 13.3 straightens that curve out by taking logarithms, which is why the log plot is the one used for measuring \(\lambda\).
13.5.6 Units of Radioactivity
The rate of radioactive decay is expressed in dps. The unit for radioactivity is the curie (Ci).
\[1\ \text{Ci} = 3.7\times10^{10}\ \text{dps (disintegrations per second)}\]
Another unit of radioactivity is the becquerel (Bq), which is simply one disintegration per second:
\[1\ \text{Bq} = 1\ \text{dps}\]
The two are therefore related by
\[1\ \text{Ci} = 3.7\times10^{10}\ \text{dps} = 3.7\times10^{10}\ \text{Bq}\]
Because 1 Bq is so small a quantity of activity, practical sources are often quoted in kBq or MBq, while the curie — being enormous — is usually met as the millicurie (mCi) or microcurie (\(\mu\)Ci).
With the mathematics of decay in place, we can now ask what particle actually leaves the nucleus in each case, and what it does to the mass number and atomic number of the parent. That is the subject of the modes of decay.
13.6 Modes of Decay
Radioelements decay by three ways, namely \(\alpha\)-decay, \(\beta\)-decay and \(\gamma\)-emission. Each of the three does something different to the nucleus, and the whole of this section can be reduced to one question asked three times: what happens to the mass number \(A\) and to the atomic number \(Z\)?
Which route a particular nuclide takes is not arbitrary. It depends on why the nucleus was unstable in the first place, as summarised in FC 13.6.
Flowchart
FC 13.6 – Choosing a mode of decay: unstable nucleus → branch 1: very heavy nuclide (A greater than about 209, too many protons for the short-range attraction to hold) → α-decay → A falls by 4, Z falls by 2 → daughter two places to the left in the periodic table; branch 2: nuclide lies above the stability belt (too many neutrons for its proton count) → β⁻-decay → A unchanged, Z rises by 1 → daughter one place to the right; branch 3: daughter nucleus left in an excited state after α- or β-decay → γ-emission → A and Z both unchanged → daughter drops to its ground state
13.6.1 Alpha Decay
The emission of an \(\alpha\)-particle from the nuclei is called \(\alpha\)-decay. You have learnt that the charge of an \(\alpha\)-particle is \(+2\) and its mass is 4 u. It is identical with a helium nucleus, and hence an \(\alpha\)-particle is also designated as \(^{4}_{2}\mathrm{He}\).
In the \(\alpha\)-decay process, the parent nucleus \(^{A}_{Z}\mathrm{X}\) emits an \(\alpha\)-particle and produces a daughter nucleus Y. The parent nucleus thus loses two protons (charge \(+2\)) and two neutrons. The total mass lost is 4 u.
The daughter nucleus will therefore have a mass 4 units less and a charge 2 units less than its parent. Book-keeping is all that is needed here, and it is worth doing slowly the first time: two protons leaving reduces \(Z\) by 2, and two protons plus two neutrons leaving reduces \(A\) by 4. The \(\alpha\)-decay process is then written as the general equation
\[^{A}_{Z}\mathrm{X} \longrightarrow\ ^{A-4}_{Z-2}\mathrm{Y} + ^{4}_{2}\mathrm{He}\]
where \(^{A}_{Z}\mathrm{X}\) is the parent, \(^{A-4}_{Z-2}\mathrm{Y}\) is the daughter and \(^{4}_{2}\mathrm{He}\) is the emitted particle. The loss of two protons and two neutrons is drawn in Fig. 13.6a.
Diagram
Fig. 13.6a – Parent nucleus drawn as a cluster of filled circles (protons) and open circles (neutrons) labelled “ᴬ_Z X”; an arrow to the right showing a compact group of 2 protons + 2 neutrons leaving, labelled “α-particle, ⁴₂He”; the remaining smaller cluster on the right labelled “daughter, ᴬ⁻⁴_(Z−2) Y”. Annotate “A decreases by 4” and “Z decreases by 2”.
For example, radium-226 decays to form radon-222:
\[^{226}_{88}\mathrm{Ra} \longrightarrow\ ^{222}_{86}\mathrm{Rn} + ^{4}_{2}\mathrm{He}\]
Uranium-238 decays to give thorium-234:
\[^{238}_{92}\mathrm{U} \longrightarrow\ ^{234}_{90}\mathrm{Th} + ^{4}_{2}\mathrm{He}\]
Note that in \(\alpha\)-decay, the daughter nucleus formed belongs to an element that occupies two places to the left in the periodic table. This follows directly from \(Z\) falling by 2, since position in the periodic table is fixed by the atomic number.
13.6.2 \(\beta\)-Decay
The emission of a negatively charged stream of \(\beta^-\) particles from the nucleus is called \(\beta^-\)-decay. You know that \(\beta^-\)-particles are electrons, with the charge and the mass of an electron, the mass being negligible as compared to the nuclei.
A nucleus decays by emitting a high-speed electron called a beta particle (\(\beta\)). A new daughter nucleus is formed with the same mass number as the original parent nucleus, but an atomic number that is one unit greater.
\[^{A}_{Z}\mathrm{X} \longrightarrow\ ^{A}_{Z+1}\mathrm{Y} + ^{0}_{-1}\mathrm{e}\]
Here \(^{A}_{Z}\mathrm{X}\) is the parent, \(^{A}_{Z+1}\mathrm{Y}\) is the daughter and \(^{0}_{-1}\mathrm{e}\) is the emitted particle. Note that the mass number \(A\) does not change; the atomic number changes.
The superscript and subscript on \(^{0}_{-1}\mathrm{e}\) are worth reading carefully, because they explain the whole result. The 0 records that an electron contributes nothing to the nucleon count, so \(A\) cannot change. The \(-1\) records its charge, and since total charge must balance, the daughter’s \(Z\) must rise by 1 to compensate.
Examples of beta decay follow. Neptunium-238 decays to form plutonium-238:
\[^{238}_{93}\mathrm{Np} \longrightarrow\ ^{238}_{94}\mathrm{Pu} + ^{0}_{-1}\mathrm{e}\]
Plutonium-241 decays to form americium-241:
\[^{241}_{94}\mathrm{Pu} \longrightarrow\ ^{241}_{95}\mathrm{Am} + ^{0}_{-1}\mathrm{e}\]
In both examples the mass number on the left equals the mass number on the right, while the atomic number has moved up by one — the daughter lies one place to the right in the periodic table.
13.6.3 \(\gamma\)-Decay
\(\gamma\)-Radiation is almost always accompanied with \(\alpha\) and \(\beta^-\) decay processes. In the \(\alpha\) and \(\beta^-\) decay process, the daughter nucleus formed is in an energetically excited state. Ground state products are formed with the emission of \(\gamma\)-rays.
For example,
\[^{238}_{92}\mathrm{U} \longrightarrow\ ^{234}_{90}\mathrm{Th} + ^{4}_{2}\mathrm{He} + \gamma\]
\(^{238}_{92}\mathrm{U}\) emits \(\alpha\)-particles of two different energies, 4.147 MeV (23%) and 4.195 MeV (77%). When \(\alpha\)-particles of energy 4.147 MeV are emitted, \(^{234}\mathrm{Th}\) is left in an excited state, which de-excites to the ground state with the emission of \(\gamma\)-ray photons of energy 0.048 MeV.
These three numbers fit together exactly, and the arithmetic is the point of the example. The two \(\alpha\)-energies differ by \(4.195 - 4.147 = 0.048\) MeV, which is precisely the energy of the \(\gamma\)-photon. When the more energetic \(\alpha\)-particle is emitted, the thorium nucleus is left in its ground state and no \(\gamma\)-ray follows; when the less energetic one is emitted, the 0.048 MeV that the \(\alpha\)-particle did not carry away stays behind in the nucleus as excitation energy, and is released a moment later as a \(\gamma\)-photon. Either way the total energy given out is the same, as shown in Fig. 13.6b.
Diagram
Fig. 13.6b – Nuclear energy-level diagram: top level labelled “²³⁸₉₂U”; two downward arrows to two levels of ²³⁴Th — the longer arrow labelled “α, 4.195 MeV (77%)” ending on the lower line labelled “²³⁴Th ground state”, the shorter arrow labelled “α, 4.147 MeV (23%)” ending on a line just above it labelled “²³⁴Th excited state”; a short vertical arrow from the excited line down to the ground line labelled “γ photon, 0.048 MeV”.
Since \(\gamma\)-emission changes neither \(Z\) nor \(A\), the nucleus before and after emission is the same nuclide in two different energy states — precisely the relationship met in section 13.2.1.
All three modes so far are spontaneous: the nucleus decides, and we only watch. The next section takes up nuclear changes that are deliberately brought about.
13.7 Nuclear Reactions
We have so far discussed natural (spontaneous) nuclear reactions through the \(\alpha\)- and \(\beta\)-decay processes. In every one of them the nucleus acted on its own, and all we could do was measure how fast it did so.
Now we consider the non-spontaneous (man-made) nuclear reactions, or nuclear transmutations. Here the nucleus is made to change by being struck with a particle from outside, which means that for the first time the chemist, and not the nucleus, decides what will happen.
13.7.1 Transmutation
The nuclear transmutation is the transformation of a stable nucleus into another nucleus, be it stable or unstable. The nuclear transmutation where the product nucleus is radioactive is called artificial radioactivity.
Before going further it is worth setting nuclear reactions side by side with the chemical reactions you have studied for two years, because almost every habit of thought carried over from chemical reactions has to be revised. The differences are collected in Table 13.3.
Table 13.3 : Comparison of chemical reactions and nuclear reactions
| Chemical Reactions | Nuclear Reactions |
|---|---|
| 1. Rearrangement of atoms by breaking and forming of chemical bonds. | 1. Elements or isotopes of one element are converted into another element in a nuclear reaction. |
| 2. Different isotopes of an element have same behaviour. | 2. Isotopes of an element behave differently. |
| 3. Only outer shell electrons take part in the chemical reaction. | 3. In addition to electrons, protons, neutrons, other elementary particles may be involved. |
| 4. The chemical reaction is accompanied by relatively small amounts of energy. e.g. chemical combustion of 1.0 g methane releases only 56 kJ energy. | 4. The nuclear reaction is accompanied by a large amount of energy change. e.g. The nuclear transformation of 1 g of Uranium-235 releases \(8.2 \times 10^{7}\) kJ |
| 5. The rates of reaction is influenced by the temperature, pressure, concentration and catalyst. | 5. The rate of nuclear reactions are unaffected by temperature, pressure and catalyst. |
Row 4 deserves a second look, because the two figures in it are the reason nuclear energy matters at all. Burning one gram of methane and transmuting one gram of uranium-235 differ in energy yield by a factor of well over a million, and that gap is simply the difference between rearranging electrons and rearranging nucleons.
13.7.2 Induced or Artificial Radioactivity
It is a type of nuclear transmutation in which the stable nucleus is converted into a radioactive nucleus. The product nucleus decays spontaneously with the emission of radiation and particles.
For example, the stable element \(^{10}\mathrm{B}\), when bombarded with \(\alpha\)-particles, transforms into \(^{13}\mathrm{N}\), which spontaneously emits positrons.
\[^{10}_{5}\mathrm{B} + ^{4}_{2}\mathrm{He} \longrightarrow\ ^{13}_{7}\mathrm{N} + ^{1}_{0}\mathrm{n}\]
\[^{13}_{7}\mathrm{N} \longrightarrow\ ^{13}_{6}\mathrm{C} + ^{0}_{1}\mathrm{e}\]
The boron nuclide on the left of the first equation is stable; the nitrogen-13 produced is radioactive, and it disintegrates by the spontaneous emission of a positron. A positron is written \(^{0}_{1}\mathrm{e}\) — a particle with the mass of an electron but a positive charge, so it leaves the mass number unchanged while reducing the atomic number by one unit, which is why nitrogen (\(Z = 7\)) becomes carbon (\(Z = 6\)).
The two-stage character of the whole process is what the term “induced radioactivity” records, and it is set out in FC 13.7a.
Flowchart
FC 13.7a – Induced radioactivity in two stages: stable target nucleus (¹⁰B) → bombardment with α-particle (forced, non-spontaneous) → transmutation → radioactive product nucleus (¹³N) + neutron → stage 2: spontaneous decay, needing no bombardment → positron emission → final stable nuclide (¹³C)
13.7.3 Nuclear Fission
Nuclear fission is the splitting of the heavy nucleus of an atom into two nearly equal fragments, accompanied by the release of a large amount of energy. When a uranium nucleus absorbs a neutron, it breaks and releases energy (heat), more neutrons, and other radiation.
For example, a \(^{235}\mathrm{U}\) nucleus captures neutrons and splits into two lighter fragments, as shown in Fig. 13.7a.
\[^{235}_{92}\mathrm{U} + ^{1}_{0}\mathrm{n} \longrightarrow\ ^{142}_{56}\mathrm{Ba} + ^{91}_{36}\mathrm{Kr} + 3\,^{1}_{0}\mathrm{n} + \text{Energy}\]
Diagram
Fig. 13.7a – Fission event drawn left to right: a slow neutron approaching a large ²³⁵U nucleus; the nucleus shown absorbing it and becoming distorted; then splitting into two unequal fragments labelled ¹⁴²₅₆Ba and ⁹¹₃₆Kr, with three neutrons flying off and a burst labelled “energy ≈ 200 MeV”. Check-mark note: mass numbers 235 + 1 = 142 + 91 + 3.
Each fission may lead to different products. The mass of the fission products is less than the parent nucleus, and a large amount of energy corresponding to the mass loss is released in each fission — the mass defect idea of section 13.3.5 applied to a splitting rather than to a formation.
When one uranium-235 nucleus undergoes fission, three neutrons are emitted, which subsequently disintegrate three more uranium nuclei and thereby produce nine neutrons. Such a chain continues by itself. In a very short time an enormous amount of energy is liberated, which can be utilised for destructive purposes (dangerous explosives) or peaceful purposes (nuclear reactor — power plant).
There is no unique way for the fission of \(^{235}\mathrm{U}\) that produces Ba and Kr, and 400 ways for the fission of \(^{235}\mathrm{U}\) leading to 800 fission products are known. Many of these are radioactive, and they undergo spontaneous disintegrations giving rise to new elements in the periodic table. The equation above is therefore one representative example, not the fission reaction.
13.7.4 Chain Reaction
A single fission is a one-off event. What makes fission an energy source rather than a laboratory curiosity is that the neutrons it produces can set off further fissions.
Interestingly, each fission emits 2 to 3 neutrons (on an average 2.5 neutrons). Can you imagine what will happen due to this neutron emission?
Obviously these neutrons emitted in fission cause more fission of the uranium nuclei, which yield more neutrons. These neutrons again bring forth fission, producing further neutrons. The process continues indefinitely, leading to a chain reaction, which continues even after the removal of the bombarding neutrons. Energy released per fission is \(\sim 200\) MeV. The multiplication is shown in Fig. 13.7b.
Diagram
Fig. 13.7b – Branching tree of a chain reaction: one neutron striking one ²³⁵U nucleus at the top; three neutrons emerging and striking three ²³⁵U nuclei in the second row; nine neutrons emerging and striking nine nuclei in the third row; dots indicating continuation. Label the rows “1st generation”, “2nd generation”, “3rd generation” and annotate “each fission releases ≈ 200 MeV”.
The word to notice is “self-propagating”. An ordinary reaction stops when a reactant is used up or when heating is discontinued; a chain reaction supplies its own trigger at every step, which is what makes it both powerful and, unless deliberately controlled, dangerous.
13.7.5 Nuclear Fusion
Energy received by the earth from the Sun is the result of nuclear fusion. In the process, the lighter nuclei combine (fuse) together and form a heavy nucleus, which is accompanied by an enormous amount of energy.
Representative fusion reactions occurring in the Sun and stars are:
\[^{1}_{1}\mathrm{H} + ^{1}_{1}\mathrm{H} \longrightarrow\ ^{2}_{1}\mathrm{H} + ^{0}_{1}\mathrm{e}\]
\[^{1}_{1}\mathrm{H} + ^{2}_{1}\mathrm{H} \longrightarrow\ ^{3}_{2}\mathrm{He}\]
\[^{3}_{2}\mathrm{He} + ^{3}_{2}\mathrm{He} \longrightarrow\ ^{4}_{2}\mathrm{He} + 2\,^{1}_{1}\mathrm{H}\]
\[^{3}_{2}\mathrm{He} + ^{1}_{1}\mathrm{H} \longrightarrow\ ^{4}_{2}\mathrm{He} + ^{0}_{1}\mathrm{e}\]
Read as a sequence, these four steps show hydrogen being built up into helium: two protons give the heavy hydrogen nuclide, that combines with another proton to give helium-3, and helium-3 goes on to helium-4. The route is summarised in FC 13.7b.
Flowchart
FC 13.7b – Fusion route in the Sun and stars: ¹₁H + ¹₁H → ²₁H + positron → ²₁H + ¹₁H → ³₂He → branch 1: ³₂He + ³₂He → ⁴₂He + 2 ¹₁H (protons returned to the start of the cycle) → branch 2: ³₂He + ¹₁H → ⁴₂He + positron; net outcome: light hydrogen nuclei built up into ⁴₂He with release of enormous energy
Advantages and Difficulty of Fusion
Nuclear fusion has certain advantages over nuclear fission. Fusion produces relatively more energy per given mass of fuel.
The only difficulty with fusion is that it requires an extremely high temperature, typically \(10^{8}\ \text{K}\). Such a temperature is needed because two nuclei approaching each other are both positively charged, and they must be driven close enough for the short-range nuclear attraction of section 13.3.4 to take over from the coulombic repulsion pushing them apart. Only at these temperatures do nuclei move fast enough to achieve that.
Fission and fusion release energy on a scale nothing chemical can match. The final section of the chapter looks at what this, and the decay processes met earlier, are actually put to work doing.
13.8 Applications of Radioisotopes
13.8.1 Radiocarbon Dating
The technique is used to find the age of historic and archaeological organic samples such as old wood samples and animal or human fossils. What makes it possible is that one carbon nuclide is radioactive, and that living things keep taking carbon in while they are alive and stop the moment they die.
The reasoning proceeds in six steps, and the whole sequence is collected in FC 13.8a.
Flowchart
FC 13.8a – Radiocarbon dating: cosmic-ray neutrons strike ¹⁴N in the upper atmosphere → ¹⁴C formed → ¹⁴C + O₂ → ¹⁴CO₂, mixes with ordinary ¹²CO₂ → absorbed by plants in photosynthesis → animals eat plants → ¹⁴C/¹²C ratio in the living body stays constant → organism dies → uptake stops, ¹⁴C decays (t½ = 5730 y) → ¹⁴C/¹²C ratio falls → measure activity N of the sample and N₀ of fresh material → age from t = (2.303/λ) log₁₀(N₀/N)
i. Radioactive \(^{14}\mathrm{C}\) is formed in the upper atmosphere by the bombardment of neutrons from cosmic rays on \(^{14}\mathrm{N}\).
\[^{14}_{7}\mathrm{N} + ^{1}_{0}\mathrm{n} \longrightarrow\ ^{14}_{6}\mathrm{C} + ^{1}_{1}\mathrm{H}\]
ii. \(^{14}\mathrm{C}\) combines with atmospheric oxygen to form \(^{14}\mathrm{CO_2}\), which mixes with ordinary \(^{12}\mathrm{CO_2}\).
iii. Carbon dioxide is absorbed by plants during photosynthesis. Animals eat plants. Hence \(^{14}\mathrm{C}\) becomes a part of plant and animal bodies.
iv. As long as the plant is alive, the ratio \(^{14}\mathrm{C}/^{12}\mathrm{C}\) remains constant. When the plant dies, photosynthesis occurs no more, and the ratio \(^{14}\mathrm{C}/^{12}\mathrm{C}\) decreases with the decay of \(^{14}\mathrm{C}\) (half life 5730 years).
\[^{14}_{6}\mathrm{C} \longrightarrow\ ^{14}_{7}\mathrm{N} + ^{0}_{-1}\mathrm{e}\]
Two opposing processes are at work while the plant lives — fresh \(^{14}\mathrm{C}\) arriving through photosynthesis and old \(^{14}\mathrm{C}\) decaying away — and they balance, which is why the ratio holds steady. Death removes only the first of the two, and from that moment the ratio can only fall.
v. The activity (\(N\)) of the given wood sample and that of a fresh sample of live plant (\(N_0\)) are measured. \(N_0\) denotes the activity of the given sample at the time of death.
vi. The age of the given wood sample — rather, the period over which it has remained dead — can be determined:
\[t = \frac{2.303}{\lambda}\log_{10}\frac{N_0}{N}\]
where \(\lambda = \dfrac{0.693}{5730\ \text{y}} = 1.21 \times 10^{-4}\ \text{y}^{-1}\).
This is nothing new mathematically: it is Eq. (13.9) of section 13.5.3 rearranged to give \(t\) instead of \(\lambda\). The only new idea in dating is what \(N_0\) and \(N\) are taken to be.
13.8.2 Electrical Energy from Nuclear Fission
Do you know what nuclear power is? It is simply the electricity made from the fission of uranium and plutonium.
Why Should We Care About Nuclear Power?
a. It offers huge environmental benefits in producing electricity.
b. It releases zero carbon dioxide.
c. It releases zero sulfur and nitrogen oxides. These are atmospheric pollutants which pollute the air. Thus it is a clean energy.
The reason for points b and c is worth stating plainly: nothing is being burnt. A fission reactor takes its energy from the nucleus, so no carbon is oxidised and no sulfur or nitrogen impurities are sent up a chimney.
We Are Interested in Nuclear Power Because
Fission of 1 gram of uranium-235 produces about 24,000 kW/h of energy. This is the same as burning 3 tons of coal, or 12 barrels of oil, or nearly 5000 m\(^3\) of natural gas.
Nuclear fission is an alternative energy source. Traditionally, the steam that drives a power station comes from boilers fuelled by oil, gas or coal. These sources are depleting, and the increasing costs of petroleum suggest we need to depend on nuclear fission energy for electricity.
Nuclear Reactor
A nuclear reactor is a device for using atomic energy in a controlled manner for peaceful purposes. During nuclear fission energy is released, and the released energy can be utilised to generate electricity in a nuclear reactor.
The word to underline is controlled. Section 13.7.3 described a chain reaction that multiplies without limit; a reactor is the arrangement that keeps the same reaction going at a steady rate instead. The set-up is shown in Fig. 13.4 (a).
Diagram
Fig. 13.4 (a) – Simplified nuclear reactor in section: outer casing labelled “radiation shield”; inside, vertical uranium rods immersed in H₂O; a neutron source arrow labelled “source to start reaction”; bubbles rising to an outlet labelled “high pressure steam”; a water inlet at the bottom labelled “water”.
In a nuclear reactor, \(\mathrm{U}^{235}\) or \(\mathrm{Pu}^{239}\), a fissionable material, is stacked with heavy water (\(\mathrm{D_2O}\), deuterium oxide) or graphite, called the moderator. The neutrons produced in the fission pass through the moderator and lose a part of their energy. Subsequently the slow neutrons are captured, which initiate new fission.
Cadmium rods are inserted in the moderator, and they have the ability to absorb neutrons. The rate of the chain reaction is thus controlled. Pushing the rods further in removes more neutrons from circulation and slows the reaction; withdrawing them lets more neutrons survive to cause fission, and the rate rises.
The energy released in such a reaction appears as heat, and is removed by circulating a liquid (coolant). The coolant, which has absorbed the excess heat from the reactor, is passed over a heat exchanger for producing steam, which is then passed through the turbines to produce electricity. The full arrangement is drawn in Fig. 13.4 (b) and summarised as a sequence in FC 13.8b.
Diagram
Fig. 13.4 (b) – Schematic of a nuclear power plant: nuclear reactor vessel with uranium fuel and control rods at the left, liquid water under high pressure carrying heat to the steam generator; steam generator feeding steam to a turbine; turbine coupled to a generator; condenser (steam from turbine condensed by river water, marked 100 °F in, 80 °F out); cooling tower; pump returning water to the reactor. Label each unit.
Flowchart
FC 13.8b – From fission to electricity: fissionable U-235 or Pu-239 fuel rods → fission releases fast neutrons + heat → moderator (D₂O or graphite) slows the neutrons → slow neutrons captured → new fission (chain reaction sustained) → cadmium control rods absorb surplus neutrons → rate held steady → heat carried away by circulating coolant → coolant passes over heat exchanger → steam produced → steam drives turbines → electricity generated
Thus the atomic energy produced with the use of the fission reaction can be controlled in the nuclear reactor and explored for peaceful purposes, such as conversion into electrical energy generating power for civilian purposes, ships, submarines and so on.
13.8.3 Applications in Medicine
Hospitals and larger medical clinics typically have a Department of Nuclear Medicine. Several radioisotopes are used for the treatment of fatal diseases like cancer.
Table 13.4 (a) : Diagnostic Radioisotopes
| Isotope | Half-life | Emitted particles | Uses | |
|---|---|---|---|---|
| (i) \(^{51}_{24}\mathrm{Cr}\) | Chromium-51 | 28 days | gamma | spleen imaging |
| (ii) \(^{59}_{26}\mathrm{Fe}\) | Iron-59 | 45 days | beta, gamma | bone marrow function |
Table 13.4 (b) : Therapeutic Radioisotopes
| Isotope | Half-life | Emitted particles | Uses | |
|---|---|---|---|---|
| (i) \(^{32}_{15}\mathrm{P}\) | phosphorus-32 | 14 days | beta | treatment of leukemia |
| (ii) \(^{60}_{27}\mathrm{Co}\) | cobalt-60 | 5.3 years | beta, gamma | external radiation source for cancer treatment |
| (iii) \(^{131}_{53}\mathrm{I}\) | iodine-131 | 8 days | beta, gamma | treatment of thyroid cancer |
| (iv) \(^{226}_{88}\mathrm{Ra}\) | radium-226 | 1620 years | beta, gamma | implanted in tumours |
The tables show medical uses in two different categories, namely diagnostic and therapeutic. For diagnostic purposes, short-lived isotopes are used to limit the exposure time to radiation, while therapeutic radioisotopes are used to destroy abnormal, usually cancerous cells.
Notice why iodine-131 is the choice for the thyroid in particular. The thyroid gland is the one organ of the body that concentrates iodine, so an iodine radioisotope is delivered to the target tissue by the body’s own chemistry, with no need to aim anything.
13.8.4 Other Applications of Radioisotopes
Radioisotopes find a variety of applications in diverse areas. Agricultural products are preserved by irradiating them with gamma rays from cobalt-60 or caesium-137. The radiation kills the organisms that cause spoiling and stops sprouting, and because the produce only passes through the radiation and does not retain the source, it does not itself become radioactive.
Another application is the radio-tracer technique, wherein one or more atoms of a chemical compound are replaced by a radionuclide to trace the path followed by that chemical in the system under study, by radioactivity measurement.
The technique works because of a fact established at the very start of this chapter: a radioisotope is chemically the same as the stable isotope it replaces, so it travels wherever the ordinary compound would travel, while its radiation announces where it has reached.
13.9 Summary
Nuclear chemistry is the branch of physical chemistry that deals with reactions involving changes in atomic nuclei. It began with Becquerel’s discovery of natural radioactivity, and it is used by geologists, astronomers and workers in biology and medicine.
The atom has a tiny dense nucleus of protons and neutrons (\(\sim 10^{-15}\ \text{m}\)) inside a much larger region of fast-moving electrons (\(\sim 10^{-10}\ \text{m}\)), the outer sphere being \(10^{5}\) times larger. Almost the whole mass sits in the nucleus, whose density is about \(10^{5}\) times that of ordinary matter.
A nuclide is written \(^{A}_{Z}\mathrm{X}\), where \(Z\) is the atomic number (protons) and \(A\) the mass number (nucleons), with \(A = Z + N\) and \(N = A - Z\). The nuclear radius is \(R = R_0 A^{1/3}\), with \(R_0 = 1.33 \times 10^{-15}\ \text{m}\), and since \(V \propto R^3\), \(V \propto A\).
Nuclides are classified by nucleon count as isotopes (same \(Z\)), isobars (same \(A\)), mirror nuclei (isobars with \(Z\) and \(N\) interchanged, differing by 1), isotones (same \(N\)) and nuclear isomers (same \(Z\) and \(N\), different energy states, the higher one marked m). By stability they are either stable or unstable (radioactive).
Four factors govern nuclear stability: nuclides with even \(Z\) and even \(N\) are the most stable (165 of them, 85% of the crust) while odd–odd nuclides are least stable (only four); light nuclides are stable at \(N/Z \approx 1\) and heavier ones need \(N/Z > 1\), all stable nuclides lying in the stability zone of Fig. 13.1; the magic numbers 2, 8, 20, 28, 50, 82 and 126 give special stability, as in \(^{208}_{82}\mathrm{Pb}\); and the charge-independent, short-range p–p, n–n and p–n attractions make up the nuclear potential that holds the nucleus together against coulombic repulsion.
The actual mass of a nuclide is less than the sum of the masses of its constituents. This mass defect, \(\Delta m\) = calculated mass − observed mass, appears as the nuclear binding energy, the energy required to break the nucleus into its constituents:
\[\Delta m = [Z\,m_H + (A - Z)\,m_n] - m \qquad \Delta E = \Delta m \times c^2\]
Nuclear masses are expressed in the unified mass unit (\(1\ \text{u} = 1.66 \times 10^{-27}\ \text{kg}\), one-twelfth the mass of a \(^{12}\mathrm{C}\) atom) and nuclear energies in MeV (\(1\ \text{MeV} = 1.602 \times 10^{-13}\ \text{J}\)). Since 1 u corresponds to 931.4 MeV, B.E. \(= \Delta m\ (\text{u}) \times 931.4\), and the binding energy per nucleon is \(\bar{B} = \text{B.E.}/A\).
On the plot of \(\bar{B}\) against \(A\) (Fig. 13.2), light nuclides (\(A < 30\)) show peaks at multiples of 4 such as \(^{4}_{2}\mathrm{He}\), \(^{12}_{6}\mathrm{C}\) and \(^{16}_{8}\mathrm{O}\); medium nuclides (\(30 < A < 90\)) rise to a broad maximum, \(^{56}\mathrm{Fe}\) with 8.79 MeV being the most stable nuclide; heavy nuclides (\(A > 90\)) fall to 7.7 MeV near \(A \cong 210\), and beyond \(^{209}\mathrm{Bi}\), the heaviest stable nuclide, all are radioactive \(\alpha\)-emitters.
Radioactivity is the spontaneous emission of a nuclear particle and gamma radiation by an unstable nucleus, which is thereby transformed into a different nuclide. The radiations are \(\alpha\), \(\beta\) and \(\gamma\). Being a nuclear phenomenon, the probability of decay does not depend on chemical combination, temperature, pressure, a catalyst, or the age of the nucleus.
The rate of decay, or activity, is \(-\dfrac{dN}{dt}\), expressed in disintegrations per second. The rate law is
\[-\frac{dN}{dt} = \lambda N\]
where \(\lambda\), the decay constant, is the fraction of nuclei decaying in unit time.
- Integrating the rate law gives the working equations of decay:
\[\lambda = \frac{2.303}{t}\log_{10}\frac{N_0}{N} \qquad N = N_0e^{-\lambda t}\]
- The half life \(t_{1/2}\) is the time in which half the nuclei present decay, and is related to the decay constant by
\[\lambda = \frac{0.693}{t_{1/2}} \qquad \text{or} \qquad t_{1/2} = \frac{0.693}{\lambda}\]
A plot of \(\log_{10}\) (activity) against time is a straight line of slope \(-\lambda/2.303\) (Fig. 13.3), while a plot of activity against time is a falling curve (Fig. 13.4), showing that the number of radioactive atoms and hence the rate of decay both decrease with time.
Units of radioactivity: \(1\ \text{Ci} = 3.7 \times 10^{10}\) dps and \(1\ \text{Bq} = 1\) dps, so \(1\ \text{Ci} = 3.7 \times 10^{10}\ \text{Bq}\).
The three modes of decay change \(A\) and \(Z\) as follows. In \(\alpha\)-decay a \(^{4}_{2}\mathrm{He}\) nucleus is emitted, \(A\) falls by 4 and \(Z\) by 2, so the daughter lies two places to the left in the periodic table. In \(\beta^-\)-decay an electron \(^{0}_{-1}\mathrm{e}\) is emitted, \(A\) is unchanged and \(Z\) rises by 1. In \(\gamma\)-emission, which usually accompanies \(\alpha\)- and \(\beta\)-decay, an excited daughter drops to its ground state with neither \(A\) nor \(Z\) changing.
\[^{A}_{Z}\mathrm{X} \longrightarrow\ ^{A-4}_{Z-2}\mathrm{Y} + ^{4}_{2}\mathrm{He} \qquad\qquad ^{A}_{Z}\mathrm{X} \longrightarrow\ ^{A}_{Z+1}\mathrm{Y} + ^{0}_{-1}\mathrm{e}\]
Nuclear transmutation is the man-made conversion of a stable nucleus into another nucleus by bombardment; when the product is radioactive it is called artificial or induced radioactivity, as when \(^{10}_{5}\mathrm{B}\) bombarded with \(\alpha\)-particles gives \(^{13}_{7}\mathrm{N}\), which emits a positron. Nuclear reactions differ from chemical reactions in converting one element into another, in involving nucleons and other particles, in releasing far larger energies, and in being unaffected by temperature, pressure and catalysts (Table 13.3).
Nuclear fission is the splitting of a heavy nucleus into two nearly equal fragments with release of a large amount of energy, about 200 MeV per fission:
\[^{235}_{92}\mathrm{U} + ^{1}_{0}\mathrm{n} \longrightarrow\ ^{142}_{56}\mathrm{Ba} + ^{91}_{36}\mathrm{Kr} + 3\,^{1}_{0}\mathrm{n} + \text{Energy}\]
Since each fission emits 2 to 3 neutrons (on average 2.5), these neutrons cause further fissions and a self-sustaining chain reaction results, which continues even after the bombarding neutrons are removed.
Nuclear fusion is the combination of lighter nuclei into a heavier one with release of enormous energy, and is the source of the energy the earth receives from the Sun. It yields more energy per unit mass of fuel than fission, but requires temperatures of about \(10^{8}\ \text{K}\).
Radiocarbon dating uses \(^{14}\mathrm{C}\), formed in the upper atmosphere by cosmic-ray neutrons acting on \(^{14}\mathrm{N}\) and taken into living things through photosynthesis. The \(^{14}\mathrm{C}/^{12}\mathrm{C}\) ratio is constant during life and falls after death (\(t_{1/2} = 5730\) y), so the age follows from
\[t = \frac{2.303}{\lambda}\log_{10}\frac{N_0}{N}, \qquad \lambda = 1.21 \times 10^{-4}\ \text{y}^{-1}\]
In a nuclear reactor a fission chain reaction is controlled for peaceful use: \(\mathrm{U}^{235}\) or \(\mathrm{Pu}^{239}\) fuel, a moderator of heavy water or graphite to slow the neutrons, cadmium rods to absorb surplus neutrons, and a coolant that carries the heat to a heat exchanger to raise steam for the turbines. Fission of 1 g of uranium-235 yields about 24,000 kW/h, equal to burning 3 tons of coal, and releases no carbon dioxide and no sulfur or nitrogen oxides.
Radioisotopes are used in medicine as diagnostic agents, where short-lived gamma emitters such as \(^{51}_{24}\mathrm{Cr}\) and \(^{59}_{26}\mathrm{Fe}\) limit exposure, and as therapeutic agents such as \(^{32}_{15}\mathrm{P}\), \(^{60}_{27}\mathrm{Co}\), \(^{131}_{53}\mathrm{I}\) and \(^{226}_{88}\mathrm{Ra}\) that destroy cancerous cells (Table 13.4). Other uses include preserving agricultural produce by irradiation with gamma rays from cobalt-60 or caesium-137, and the radio-tracer technique, in which a radionuclide replaces an atom in a compound so that its path through a system can be followed by measuring radioactivity.
The chapter’s overall route, from the structure of the nucleus to the uses of nuclear change, is condensed in FC 13.9.
Flowchart
FC 13.9 – Chapter at a glance: nucleus described by Z and A → classification of nuclides (isotopes / isobars / mirror nuclei / isotones / nuclear isomers) → is the nuclide stable? → branch 1: stable (even–even, N/Z in the belt, magic numbers, high binding energy per nucleon) → no change; branch 2: unstable → radioactivity → rate law −dN/dt = λN → half life t½ = 0.693/λ → mode of decay: α (A−4, Z−2) / β⁻ (A same, Z+1) / γ (A and Z same) → new nuclide closer to the stability belt; separate branch: man-made change → transmutation → induced radioactivity, nuclear fission (chain reaction, ~200 MeV) and nuclear fusion (10⁸ K) → applications: radiocarbon dating, electricity from a controlled reactor, diagnosis and therapy in medicine, food preservation and radio-tracers
13.10 Check Your Concepts
Assertion (A): The nucleus accounts for almost the whole mass of an atom but for almost none of its volume. Reason (R): The mass of an electron is about \(1/1837\) that of a proton, and the nuclear radius is of the order of \(10^{-15}\) m against about \(10^{-10}\) m for the atom. State whether R is a correct explanation of A.
The constant \(R_0\) in \(R = R_0A^{1/3}\) has the same value for every nucleus. What does this single fact tell you about how tightly nucleons are packed in light and heavy nuclei alike?
Distinguish between isobars and isotones, giving one example of each. Then explain why every pair of mirror nuclei is also a pair of isobars, while most isobars are not mirror nuclei.
\(^{60m}\mathrm{Co}\) and \(^{60}\mathrm{Co}\) have the same number of protons and the same number of neutrons. In what respect do they differ, what is the higher-energy form called, and how is it indicated in the symbol?
Match each pair in Column I with the correct relationship in Column II.
| Column I | Column II |
|---|---|
| a. \(^{22}_{11}\mathrm{Na}\) and \(^{24}_{11}\mathrm{Na}\) | i. isobars |
| b. \(^{14}_{6}\mathrm{C}\) and \(^{14}_{7}\mathrm{N}\) | ii. isotones |
| c. \(^{13}_{6}\mathrm{C}\) and \(^{14}_{7}\mathrm{N}\) | iii. mirror nuclei |
| d. \(^{13}_{6}\mathrm{C}\) and \(^{13}_{7}\mathrm{N}\) | iv. isotopes |
Only four stable naturally occurring nuclides have both \(Z\) and \(N\) odd. Explain this observation in terms of nucleon pairing, and state why a leftover proton cannot pair with a leftover neutron.
Explain why the belt of stable nuclides follows the line \(N = Z\) for light nuclides but bends upward, away from that line, as \(Z\) increases.
What are magic numbers? Why is \(^{208}_{82}\mathrm{Pb}\) described as having two magic numbers, and what does this predict about its stability?
If coulombic repulsion were the only force acting inside a nucleus, no nucleus could exist. What conclusion is drawn from the fact that nuclei do exist, and which three attractions make up the nuclear potential?
Why is the binding energy per nucleon, rather than the total binding energy, used when the stabilities of two nuclides are compared?
Arrange \(^{2}\mathrm{H}\), \(^{56}\mathrm{Fe}\) and \(^{238}\mathrm{U}\) in increasing order of binding energy per nucleon, and justify the order using the shape of the curve in Fig. 13.2.
Assertion (A): Radioactive decay follows a first-order rate law. Reason (R): The rate of decay at any instant is proportional to the number of nuclei present at that instant. State whether R is a correct explanation of A.
Why is a negative sign written in the expression for the rate of decay, \(-\dfrac{dN}{dt}\), when the rate itself is a positive quantity?
Explain why a radioactive sample never decays completely, and why chemists therefore quote a half life rather than a time for total decay.
What happens to the half life of a radioelement if (a) the sample is heated strongly, (b) the element is converted into one of its compounds, (c) the mass of the sample is doubled? Give a reason for your answer in each case.
Complete and balance each nuclear equation, and name the type of change taking place:
- \(^{226}_{88}\mathrm{Ra} \longrightarrow\ ? + ^{4}_{2}\mathrm{He}\)
- \(^{214}_{82}\mathrm{Pb} \longrightarrow\ ^{214}_{83}\mathrm{Bi} + \ ?\)
- \(^{235}_{92}\mathrm{U} + ^{1}_{0}\mathrm{n} \longrightarrow\ ^{140}_{54}\mathrm{Xe} + \ ? + 2\,^{1}_{0}\mathrm{n}\)
A nuclide lies below the stability belt, that is, it has too few neutrons for its proton count. Using the changes in \(N\) and \(Z\), explain which of \(\beta^-\) emission or positron emission would bring it closer to the belt.
Why does \(\gamma\)-emission almost always accompany \(\alpha\)- and \(\beta\)-decay, and why does it leave the identity of the element unchanged?
\(^{238}_{92}\mathrm{U}\) emits \(\alpha\)-particles of two different energies, 4.147 MeV and 4.195 MeV, and a \(\gamma\)-photon of 0.048 MeV is observed only with the first of these. Explain this observation.
State two differences between nuclear fission and nuclear fusion. Explain why fusion requires a temperature of about \(10^{8}\) K while fission does not require high temperature at all.
Explain why a chain reaction continues even after the bombarding neutrons are withdrawn, and what feature of the fission of \(^{235}\mathrm{U}\) makes this possible.
In a nuclear reactor, state and explain the effect on the chain reaction of (a) removing the moderator, (b) pushing the cadmium rods further into the core, (c) stopping the circulation of the coolant.
Isotopes of an element are chemically indistinguishable, yet only some of them are radioactive. Explain this, and state how the first half of the statement makes the radio-tracer technique possible.
Explain why short-lived isotopes are preferred for diagnostic work while longer-lived ones are used in therapy, giving one example of each from Table 13.4.
Why can radiocarbon dating not be applied to a stone tool or to a rock, and what quantity must be measured for the method to be used at all?
Assertion (A): The rate of a nuclear reaction is not altered by a catalyst. Reason (R): A catalyst acts on the outer electrons of the reacting atoms, whereas a nuclear reaction is a change in the nucleus. State whether R is a correct explanation of A.
On the basis of even-odd nature of protons and neutrons, what type of nuclides are most stable?
Explain in brief, nuclear fission.
The nuclides with an odd number of both protons and neutrons are the least stable. Why?
Referring to the stability belt of stable nuclides, which nuclides are \(\beta^-\) and \(\beta^+\) emitters? Why?
Explain with an example each nuclear transmutation and artificial radioactivity. What is the difference between them?
What is binding energy per nucleon? Explain with the help of a diagram how binding energy per nucleon affects nuclear stability.
Explain with an example \(\alpha\)-decay.
Energy produced in nuclear fusion is much larger than that produced in nuclear fission. Why is it difficult to use fusion to produce energy?
How does the \(N/Z\) ratio affect nuclear stability? Explain with a suitable diagram.
You are given a very old sample of wood. How will you determine its age?
Give an example of mirror nuclei.
Balance the nuclear reaction:
\[^{118}_{54}\mathrm{Xe} \longrightarrow\ ? + ^{118}_{53}\mathrm{I}\]
Name the most stable nuclide known. Write two factors responsible for its stability.
Write the relation between the decay constant of a radioelement and its half life.
What is the difference between an \(\alpha\)-particle and a helium atom?
Write one point that differentiates nuclear reactions from chemical reactions.
Write pairs of isotones and one pair of mirror nuclei from the following:
\[^{10}_{5}\mathrm{B},\quad ^{12}_{6}\mathrm{C},\quad ^{27}_{13}\mathrm{Al},\quad ^{11}_{6}\mathrm{C},\quad ^{28}_{14}\mathrm{S}\]
Derive the relationship between half life and decay constant of a radioelement.
Represent graphically \(\log_{10}\) (activity / dps) versus \(t\)/s. What is its slope?
Write two units of radioactivity. How are they interrelated?
Write the fusion reactions occurring in the Sun and stars.
A produces B by \(\alpha\)-emission. If B is in group 16 of the periodic table, what is the group of A?
13.11 Practice With Numericals
For the nuclide \(^{238}_{92}\mathrm{U}\), state the number of protons, neutrons and nucleons, and calculate its \(N/Z\) ratio.
Calculate the radius of the \(^{64}_{29}\mathrm{Cu}\) nucleus, taking \(R_0 = 1.33 \times 10^{-15}\ \text{m}\). What is the ratio of this radius to that of a nucleus of mass number 8?
The atomic mass of \(^{14}_{7}\mathrm{N}\) is 14.0031 u. Taking the mass of the H atom as 1.0078 u and that of the neutron as 1.0087 u, calculate (a) the mass defect, (b) the total binding energy in MeV, and (c) the binding energy per nucleon.
The total binding energy of \(^{40}_{20}\mathrm{Ca}\) is 342 MeV. Calculate its binding energy per nucleon, and state whether the value places it to the left of, at, or to the right of the maximum in Fig. 13.2.
Calculate the energy equivalent of a mass defect of 0.0500 u, expressing the answer in MeV and in joules. (Take \(1\ \text{u} \equiv 931.4\ \text{MeV}\) and \(1\ \text{MeV} = 1.602 \times 10^{-13}\ \text{J}\).)
The half life of \(^{59}_{26}\mathrm{Fe}\) is 45 days. Calculate its decay constant in \(\text{d}^{-1}\).
The decay constant of a radioisotope is \(4.5 \times 10^{-2}\ \text{h}^{-1}\). Calculate its half life in hours.
A radioisotope has a half life of 12 hours. What fraction of a given sample will remain undecayed after 2 days, and what percentage will have decayed in that time?
The decay constant of a radioelement is \(0.023\ \text{d}^{-1}\). Calculate the time required for 90% of a given sample to decay.
Calculate the activity, in disintegrations per second and in curie, of a 2.0 mg sample of \(^{60}_{27}\mathrm{Co}\). (Half life of \(^{60}\mathrm{Co}\) = 5.3 y; take the molar mass as \(60\ \text{g mol}^{-1}\), \(N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}\) and \(1\ \text{Ci} = 3.7 \times 10^{10}\) dps.)
- Express an activity of \(3.7 \times 10^{12}\) Bq in curie. (b) Express an activity of 25 mCi in disintegrations per second.
Find the number of \(\alpha\) and \(\beta\) particles emitted in the disintegration series
\[^{238}_{92}\mathrm{U} \longrightarrow\ ^{206}_{82}\mathrm{Pb}\]
The energy released per fission of a \(^{235}\mathrm{U}\) nucleus is about 200 MeV. Calculate the total energy released, in MeV and in joules, when \(1.0 \times 10^{3}\) nuclei undergo fission.
In a chain reaction each fission releases two neutrons, and every neutron goes on to cause a fresh fission. Starting from a single fission as the first generation, calculate the number of fissions taking place in the eighth generation.
Estimate the energy released in MeV in the fusion reaction
\[^{2}_{1}\mathrm{H} + ^{2}_{1}\mathrm{H} \longrightarrow\ ^{3}_{2}\mathrm{He} + ^{1}_{0}\mathrm{n}\]
Given atomic masses: \(^{2}\mathrm{H} = 2.0141\ \text{u}\), \(^{3}\mathrm{He} = 3.0160\ \text{u}\), and mass of neutron \(= 1.0087\ \text{u}\).
A sample of old wood shows a \(^{14}\mathrm{C}\) activity of 10.0 dps per gram, while fresh wood of the same mass shows 15.0 dps per gram. Calculate the age of the sample, taking \(\lambda = 1.21 \times 10^{-4}\ \text{y}^{-1}\).
Half life of \(^{24}\mathrm{Na}\) is 900 minutes. What is its decay constant?
Decay constant of \(^{197}\mathrm{Hg}\) is \(0.017\ \text{h}^{-1}\). What is its half life?
The total binding energy of \(^{58}\mathrm{Ni}\) is 508 MeV. What is its binding energy per nucleon?
Atomic mass of \(^{32}_{16}\mathrm{S}\) is 31.97 u. If the masses of neutron and H atom are 1.0087 u and 1.0078 u respectively, what is the mass defect?
How many \(\alpha\) and \(\beta\) particles are emitted in the transmutation
\[^{232}_{90}\mathrm{Th} \longrightarrow\ ^{208}_{82}\mathrm{Pb}\ ?\]
- Find the number of \(\alpha\) and \(\beta\) particles emitted in the process
\[^{222}_{86}\mathrm{Rn} \longrightarrow\ ^{214}_{84}\mathrm{Po}\]
Half life of \(^{18}\mathrm{F}\) is 110 minutes. What fraction of an \(^{18}\mathrm{F}\) sample decays in 20 minutes?
Half life of \(^{35}\mathrm{S}\) is 87.8 d. What percentage of a \(^{35}\mathrm{S}\) sample remains after 180 d?
Half life of \(^{67}\mathrm{Ga}\) is 78 h. How long will it take to decay 12% of a sample of Ga?
A 0.5 g sample of \(^{201}\mathrm{Tl}\) decays to 0.0788 g in 8 days. What is its half life?
65% of an \(^{111}\mathrm{In}\) sample decays in 4.2 d. What is its half life?
Calculate the binding energy per nucleon of \(^{84}_{36}\mathrm{Kr}\), whose atomic mass is 83.913 u. (Mass of neutron is 1.0087 u and that of H atom is 1.0078 u.)
Calculate the energy in MeV released in the nuclear reaction
\[^{174}_{77}\mathrm{Ir} \longrightarrow\ ^{170}_{75}\mathrm{Re} + ^{4}_{2}\mathrm{He}\]
Atomic masses: Ir = 173.97 u, Re = 169.96 u and He = 4.0026 u.
Three quarters of the original amount of a radioisotope decays in 60 minutes. What is its half life?
How many \(\alpha\)-particles are emitted by 0.1 g of \(^{226}\mathrm{Ra}\) in one year?
A sample of \(^{32}\mathrm{P}\) initially shows an activity of one curie. After 303 days the activity falls to \(1.5 \times 10^{4}\) dps. What is the half life of \(^{32}\mathrm{P}\)?
Half life of radon is 3.82 d. By what time would 99.9% of radon be decayed?
It has been found that the Sun’s mass loss is \(4.34 \times 10^{9}\) kg per second. How much energy per second would be radiated into space by the Sun?
A sample of old wood shows 7.0 dps/g. If a fresh sample of tree shows 16.0 dps/g, how old is the given sample of wood? Half life of \(^{14}\mathrm{C}\) is 5730 y.
13.12 Objective Questions
- Identify the nuclear fusion reaction.
- \(^{1}_{1}\mathrm{H} + ^{1}_{1}\mathrm{H} \longrightarrow\ ^{2}_{1}\mathrm{H} + ^{0}_{1}\mathrm{e}\)
- \(^{2}_{1}\mathrm{H} + ^{1}_{1}\mathrm{H} \longrightarrow\ ^{3}_{2}\mathrm{He}\)
- \(^{3}_{1}\mathrm{H} + ^{1}_{1}\mathrm{H} \longrightarrow\ ^{3}_{1}\mathrm{H} + ^{1}_{1}\mathrm{p}\)
- The missing particle from the nuclear reaction is
\[^{27}_{13}\mathrm{Al} + ^{4}_{2}\mathrm{He} \longrightarrow\ ? + ^{1}_{0}\mathrm{n}\]
a. $^{30}_{15}\mathrm{P}$
b. $^{32}_{16}\mathrm{S}$
c. $^{14}_{10}\mathrm{Ne}$
d. $_{14}\mathrm{Si}$- \(^{60}_{27}\mathrm{Co}\) decays with a half-life of 5.27 years to produce \(^{60}_{28}\mathrm{Ni}\). What is the decay constant for such radioactive disintegration?
- \(0.132\ \text{y}^{-1}\)
- 0.138
- 29.6 y
- 13.8%
- The radioactive isotope used in the treatment of leukemia is
- \(^{60}\mathrm{Co}\)
- \(^{226}\mathrm{Ra}\)
- \(^{32}\mathrm{P}\)
- \(^{226}\mathrm{I}\)
- The process by which nuclei having low masses are united to form nuclei with large masses is
- chemical reaction
- nuclear fission
- nuclear fusion
- chain reaction