Chapter 15: Structure of Atoms and Nuclei
In 1868, astronomers travelling to observe a total solar eclipse pointed their instruments at the thin, glowing rim of gas around the darkened Sun. Instead of a smooth rainbow, the light split into a set of sharp, separated coloured lines. Such lines were already known to act like fingerprints: every element on Earth produced its own fixed set of them. But one bright yellow line in the Sun’s light matched nothing that had ever been measured in a laboratory.
The astronomers concluded that the Sun contained an element unknown on Earth. It was named helium, after the Greek word for the Sun. For nearly three decades helium existed only as a line in a spectrum — a colour with no sample, no bottle, no piece of matter to hold. Only in 1895 was helium finally isolated on Earth, from a uranium-bearing mineral.
Two puzzles were hidden in that single yellow line. Why should an atom emit only certain sharp colours and nothing in between? And why should the Sun be so rich in helium, when the mineral that finally yielded helium on Earth happened to be radioactive?
The answers lie inside the atom — in the tiny nucleus at its centre and in the electrons that surround it.
Diagram
Fig. 15.0 – A solar eclipse observed through a spectroscope, the Sun’s light split into separate bright coloured lines with one yellow line highlighted.
Each of these questions is answered somewhere in this chapter — watch for them as you read.
15.1 Introduction
An Idea That Began Without Any Experiment
The idea that matter cannot be cut endlessly is very old. The Greek philosophers Leucippus (–370 BC) and Democritus (460 – 370 BC) were the first to propose, in the 5th century BC, that matter is made of indivisible parts. They called these parts atoms.
Their claim came from reasoning alone, not from any experiment. It stayed a philosophical idea for more than two thousand years, until chemistry began measuring how substances combine.
Dalton (1766–1844) turned the idea into a scientific theory in the early nineteenth century. His atomic theory made three statements: (i) matter is made up of indestructible particles, (ii) atoms of a given element are identical, and (iii) atoms can combine with other atoms to form new substances.
Note the word “indestructible.” It means that an atom was thought to have no parts at all — nothing inside it that could be pulled out or broken off. This is exactly the point where later experiments proved Dalton wrong.
The Atom Turns Out to Have Parts
The experiments of J. J. Thomson (1856–1940), who discovered electrons in 1887, showed that atoms are not indestructible. Something smaller could be taken out of an atom, so the atom itself had to have an internal structure.
Thomson then proposed an atomic model of his own. It had some deficiencies, and it was later improved upon by Ernest Rutherford (1871–1937) and, after him, by Niels Bohr (1885–1962). Each new model kept whatever the earlier one explained correctly and repaired what it could not explain. We will discuss these different models one by one in this chapter, in the same order in which they were developed.
You have already studied atoms and nuclei in Std. XI chemistry. This chapter will enable you to consolidate those concepts and see the physics behind them.
How the Atom Is Actually Arranged
We will learn that an atom contains a tiny nucleus whose size (radius) is about 100000 times smaller than the size of the atom itself. In other words, the radius of the nucleus is about \(10^{-5}\) times the radius of the atom, as suggested in Fig. 15.1a.
Diagram
Fig. 15.1a – Scale comparison of an atom: a large sphere marking the region of the electron orbits with an extremely small dot at the centre marked “nucleus”.
The nucleus contains all the positive charge of the atom and also 99.9% of its mass. So almost everything heavy about an atom sits in an almost invisible speck at its centre, while the electrons occupy the huge space around it and decide the size of the atom.
Questions of this kind — comparing atomic and nuclear sizes, volumes or densities as powers of ten — are a common way of testing whether you have understood the scale of the atom, so practise handling such order-of-magnitude estimates carefully.
What This Chapter Will Cover
Once the structure of the atom is settled, the natural next question is what the nucleus itself is made of and how it survives. In this chapter we will therefore also study the properties of the nucleus, the forces that keep it intact, its radioactive decays, and the energy that can be obtained from it.
15.2 Thomson’s Atomic Model
The Experiment That Pulled a Particle Out of the Atom
Thomson performed several experiments with a glass vacuum tube in which a voltage was applied between two electrodes sealed inside the evacuated tube. Removing the air was essential, because in ordinary air the particles travelling between the electrodes would collide with gas molecules and never reach the far end of the tube.
When the voltage was applied, the cathode (the negative electrode) was seen to emit rays. These rays produced a glow where they struck the glass behind the anode, as shown in Fig. 15.2a. The glow acted as a detector: it showed both that something was travelling across the tube and where it landed.
Diagram
Fig. 15.2a – Evacuated glass discharge tube with cathode and anode connected to a high voltage source, rays travelling from cathode to anode and producing a glow on the glass beyond the anode.
By studying the properties of these rays, Thomson concluded that the rays are made up of negatively charged particles, which he called electrons.
This single conclusion demolished a two-thousand-year-old assumption. Atoms are not indestructible: they contain electrons, and these electrons can be emitted by the cathode. If a piece of an atom can be taken out of it, the atom must have parts inside it.
The Plum-Pudding Picture
Having found electrons inside atoms, Thomson had to explain how they sit there. He proposed his model of an atom in 1903. According to this model, an atom is a sphere having a uniform positive charge in which electrons are embedded, as shown in Fig. 15.2b. This model is referred to as the plum-pudding model, the electrons being the plums scattered through a pudding of positive charge.
Diagram
Fig. 15.2b – Thomson’s plum-pudding model: a large sphere of uniformly spread positive charge with several small electrons embedded at various points inside it.
The total positive charge is equal to the total negative charge of the electrons in the atom, rendering it electrically neutral. This is why ordinary matter shows no net charge even though it is full of charged particles.
What the Model Explained, and Where It Failed
The model also explained the formation of ions and ionic compounds. If an atom loses one or more of its embedded electrons, the remaining structure has more positive charge than negative charge and becomes a positive ion; if it gains extra electrons, it becomes a negative ion. Oppositely charged ions then attract each other electrostatically and form ionic compounds.
However, further experiments on the structure of atoms, which are described in the sections that follow, showed the distribution of charges inside the atom to be very different from what was proposed in Thomson’s model. The failure was not about neutrality or about ions, both of which the model handled well. It was about where the positive charge sits — spread thinly through the whole atom, as Thomson assumed, or packed into one tiny region. That question could only be settled by firing something at an atom and watching how it came back.
15.3 Geiger-Marsden Experiment
The Idea Behind the Experiment
In order to understand the structure of atoms, Rutherford suggested an experiment for the scattering of alpha particles by atoms. The logic was simple: if you cannot look inside an atom, throw something small and fast at it and study how it comes off. The way the projectiles are deflected then reveals how the charge and mass inside the atom are arranged.
Alpha particles are helium nuclei and are positively charged, having a charge of two protons. Being positively charged, they are pushed away by any positive charge they approach, so they are a natural probe for finding out where the positive charge of an atom is located.
The experiment was performed by Rutherford’s colleagues Geiger (1882–1945) and Marsden (1889–1970) between 1908 and 1913. A sketch of the experimental set up is shown in Fig. 15.1.
Diagram
Fig. 15.1 – Geiger-Marsden experiment: alpha source with collimator producing a narrow beam falling on a thin gold foil, a surrounding fluorescent screen, a movable microscope, and the scattering angle θ marked between the incident direction and a scattered particle.
How the Measurement Was Made
Alpha particles from a source were collimated, i.e., focused into a narrow beam, and were made to fall on a gold foil. Collimation matters, because unless the incoming particles all travel in one known direction, there is no fixed reference line from which a deflection can be measured.
The scattered particles produced scintillations on the surrounding screen. A scintillation is a tiny flash of light produced when a fast charged particle strikes a suitable material — in effect, each flash marks the arrival of one alpha particle.
The scintillations could be observed through a microscope which could be moved to cover different angles with respect to the incident beam. By parking the microscope at a chosen angle and counting flashes patiently, the experimenters could find how many alpha particles were arriving in each direction. The angle a particle is turned through, measured from the original direction of the beam, is called the scattering angle, and a typical scattering angle is shown by θ in the figure.
What the Counting Revealed
It was found that most alpha particles passed straight through the foil, while a few were deflected (scattered) through various scattering angles.
The numbers, however, are the striking part. Only about 0.14% of the incident alpha particles were scattered through angles larger than 0.1°. Even out of these, most were deflected through very small angles.
About one alpha particle in 8000 was deflected through an angle larger than 90°, and fewer still were deflected through angles as large as 180°. A deflection of 180° means the particle came straight back along the path by which it had arrived.
These fractions are worth remembering as orders of magnitude rather than as exact figures, because questions on this experiment usually ask what the smallness of the large-angle fraction tells us about the size of the scattering centre — the point taken up in the next section.
15.4 Rutherford’s Atomic Model
Why Thomson’s Model Could Not Survive the Data
The results of the Geiger-Marsden experiment could not be explained by Thomson’s model. To see why, we have to ask what that model predicts for a beam of alpha particles.
In Thomson’s model, the positive charge was uniformly spread over the large sphere constituting the atom. Spreading a fixed amount of charge over such a large volume means that the volume density of the positive charge — the charge per unit volume — would be very small everywhere inside the atom.
A thinly spread positive charge can only push weakly on a fast, heavy alpha particle passing through it. Consequently, in that picture, all of the incident alpha particles would get deflected only through very small angles. A particle turning through more than 90°, and certainly one coming straight back, is simply not possible in such a model. Yet such particles were observed.
Rutherford’s Argument
Rutherford argued that the alpha particles which were deflected back must have encountered a massive particle with a large positive charge, so that they were repelled back. Two properties are needed here: a large positive charge to produce a strong repulsion, and a large mass so that the object itself does not simply get knocked aside when the alpha particle hits it head-on.
He then used the rarity of these events as a measure of size. From the fact that an extremely small number of alpha particles turned back while most others passed through almost undeflected, he concluded that the positively charged particle in the atom must be very small in size and must contain most of the mass of the atom. If this heavy positive lump were large, a big share of the beam would have struck it; the fact that almost none did shows that it presents a very small target.
From the experimental data, the size of this particle was found to be about 10 fm (one femtometre \(= 10^{-15}\) m), which is about \(10^{-5}\) times the size of the atom. The volume of this particle was thus found to be about \(10^{-15}\) times that of an atom, since volume varies as the cube of the linear size. He called this particle the nucleus of an atom.
The Model Rutherford Proposed
He proposed that the entire positive charge and most (99.9%) of the mass of an atom is concentrated in the central nucleus, and the electrons revolve around it in circular orbits, similar to the revolution of the planets around the Sun in the Solar system. This arrangement is shown in Fig. 15.4a.
Diagram
Fig. 15.4a – Rutherford’s nuclear model of the atom: a very small dense nucleus at the centre with electrons moving along circular orbits around it, the space between orbits and nucleus marked as empty.
The revolution of the electrons was necessary, because without it the electrons would fall into the positively charged nucleus and the atom would collapse. A stationary electron placed near a nucleus feels an unbalanced attraction and is simply pulled in. If instead it moves in a circle, that same electrostatic attraction is used up in providing the centripetal force needed to keep it turning, and the electron can stay at a fixed distance from the nucleus.
How the Model Explains the Observations
The space between the orbits of the electrons — which decide the size of the atom — and the nucleus is mostly empty. This one feature accounts for the entire pattern of the scattering results.
Thus, most alpha particles pass through this empty space undeflected. A very few which are in direct line with the tiny nucleus, or are extremely close to it, get repelled and get deflected through large angles, as illustrated in Fig. 15.4b. The closer the approach, the stronger the repulsion and the larger the deflection; a nearly head-on approach sends the particle back the way it came.
Diagram
Fig. 15.4b – Several alpha particle paths approaching a nucleus: distant ones passing almost straight, a closer one bent through a moderate angle, and a head-on one turned back through nearly 180°.
This model also explains why no positively charged particles are emitted by atoms while negatively charged electrons are. The positive charge sits in the nucleus, which is extremely massive; because of this large mass, the nucleus does not get affected when a force is applied on the atom. The electrons, being light and lying far outside, are comparatively easy to dislodge.
The nuclear model, then, explains the scattering data, the size of the atom and the emission of electrons. What it does not yet explain is how such an atom can survive at all — the difficulty taken up next.
15.4.1 Difficulties with Rutherford’s Model
A Model That Is Right in Outline but Unstable in Detail
Though this model in its basic form is still accepted, it faced certain difficulties. The nucleus, the empty space and the orbiting electrons all survive into modern physics; what fails is the classical description of how those electrons move.
We know from Maxwell’s equations that an accelerated charge emits electromagnetic radiation. This is a general result of classical electromagnetic theory: whenever a charged particle’s velocity changes, it radiates energy in the form of electromagnetic waves.
An electron in Rutherford’s model moves uniformly along a circular orbit around the nucleus. Even though the magnitude of its velocity is constant, its direction changes continuously, and so the motion is an accelerated motion.
The First Difficulty: The Atom Should Collapse
Since the orbiting electron is accelerating, the electron should emit electromagnetic radiation continuously.
Radiation carries energy away, and that energy has to come from the electron itself. Also, as it emits radiation, its energy would decrease and consequently the radius of its orbit would decrease continuously. A slower, less energetic electron cannot maintain a wide orbit, so it must settle into a smaller one.
This process would not stop halfway. The electron would then spiral into the nucleus, as suggested in Fig. 15.4c, causing the atom to collapse and lose its atomic properties. An estimate based on classical theory gives a collapse time far shorter than a second, so on this picture no atom — and therefore no matter of the kind we see around us — should exist at all.
Diagram
Fig. 15.4c – An electron following an inward spiral path from its circular orbit into the nucleus, with wavy arrows marking electromagnetic radiation emitted along the way.
The Second Difficulty: The Radiation Should Be of Every Frequency
There is a second, equally serious problem, and it concerns not the atom’s survival but the light it gives out.
As the electron loses energy, its velocity changes continuously, and the frequency of the radiation emitted would also change continuously as it moves towards the nucleus. Since the orbit shrinks smoothly rather than in jumps, the emitted frequency would sweep smoothly through a whole range of values. The atom would therefore give out a continuous spread of frequencies, and it would do so all the time, without being disturbed in any way.
What Is Actually Observed
None of these things are observed.
Firstly, most atoms are very stable. They persist unchanged for enormous lengths of time, which is why matter has fixed, reliable properties at all.
Secondly, they do not constantly emit electromagnetic radiation, and definitely not radiation of varying frequency. An undisturbed atom sitting in a room is dark, not glowing.
The atoms have to be given energy, e.g., by heating, for them to be able to emit radiation. Even then, they emit electromagnetic radiations of particular frequencies, as will be seen in the next section — a small set of sharply defined frequencies, not a smoothly varying range.
Rutherford’s model failed on all these counts. It correctly located the charge and the mass, but it could not account for the stability of the atom or for the nature of the radiation atoms emit.
The two difficulties listed here — continuous emission leading to collapse, and continuously varying frequency — are frequently asked for directly in examinations, so learn them as a pair of distinct points, each with the observation that contradicts it.
15.5 Atomic Spectra
Two Very Different Kinds of Spectra
We know that when a metallic object is heated, it emits radiation of different wavelengths. When this radiation is passed through a prism, the prism bends each wavelength through a slightly different angle and spreads the light out. What we get is a continuous spectrum — an unbroken band in which every wavelength within the range is present, with no gaps.
However, the case is different when we heat hydrogen gas inside a glass tube to high temperatures. The emitted radiation has only a few selected wavelengths, and when it is passed through a prism we get what is called a line spectrum, as shown for the visible range in Fig. 15.2. Instead of a continuous band we see a few isolated bright lines separated by darkness.
Diagram
Fig. 15.2 – Hydrogen line spectrum in the visible region: a dark background with four bright lines marked 410 nm, 434 nm, 486 nm and 656 nm.
The difference between the two cases is worth stating plainly, because it is the whole clue to atomic structure. In a heated solid the atoms are packed together and interact strongly, and their combined emission fills in every wavelength. In a heated gas the atoms are far apart and radiate individually, so what we see is what a single, undisturbed atom is capable of emitting. Fig. 15.5a contrasts the two.
Diagram
Fig. 15.5a – Side-by-side comparison: continuous spectrum from a heated metal shown as an unbroken coloured band, and line spectrum from hydrogen gas shown as a few separate bright lines on a dark background.
The hydrogen spectrum shows that hydrogen emits radiations of wavelengths 410, 434, 486 and 656 nm, and does not emit any radiation with wavelengths in between these wavelengths. The lines seen in the spectrum are called emission lines.
The Lines Fall into Families
Hydrogen atom also emits radiation at some other values of wavelengths in the ultraviolet (UV), the infrared (IR) and at longer wavelengths. The four visible lines are therefore only a small part of the full hydrogen spectrum; most of it lies outside the range our eyes can detect.
The spectral lines can be divided into groups known as series, with the names of the scientists who studied them. The series, starting from shorter wavelengths and going to larger wavelengths, are called Lyman series, Balmer series, Paschen series, Brackett series, Pfund series, etc. The Lyman series lies in the ultraviolet, the Balmer series contains the four visible lines described above, and the later series lie progressively deeper in the infrared.
In each series, the separation between successive lines decreases as we go towards shorter wavelength, and they reach a limiting value. The lines therefore crowd together at the short-wavelength end of each series instead of continuing at equal spacing.
Schematic diagrams for the first three series are shown in Fig. 15.3. The limiting value of the wavelength for each series is shown by dotted lines in the figure.
Diagram
Fig. 15.3 – Schematic of the Lyman (UV), Balmer (visible) and Paschen (IR) series of hydrogen, with lines crowding together towards shorter wavelength in each series and a dotted line marking the series limit of each.
The Empirical Formula
The observed wavelengths of the emission lines are found to obey the relation
\[\frac{1}{\lambda} = R\left(\frac{1}{n^{2}} - \frac{1}{m^{2}}\right) \qquad ---(15.1)\]
Here \(\lambda\) is the wavelength of a line, measured in metres (m); \(R\) is a constant, whose SI unit is m\(^{-1}\) since the left-hand side has the unit m\(^{-1}\); and \(n\) and \(m\) are integers, which are pure numbers with no unit.
The integer \(n\) fixes the series: \(n = 1, 2, 3, \ldots\) respectively for the Lyman, Balmer, Paschen, … series. For a chosen series, \(n\) is held fixed and \(m\) takes all integral values greater than \(n\) for that series, each value of \(m\) giving one line of that series. The wavelength decreases with increase in \(m\).
Why the Lines Crowd Together
The difference in wavelengths of successive lines in each series (fixed value of \(n\)) can be calculated from Eq. (15.1) and shown to decrease with increase in \(m\). The reason is that once \(m\) is fairly large, changing it by one alters \(\dfrac{1}{m^{2}}\) only very slightly, so consecutive lines differ less and less.
Thus, the successive lines in a given series come closer and closer and ultimately reach the value of
\[\lambda = \frac{n^{2}}{R}\]
in the limit \(m \rightarrow \infty\), for different values of \(n\). This limiting wavelength is exactly the series limit marked by the dotted lines in Fig. 15.3. Setting \(\dfrac{1}{m^{2}} = 0\) in Eq. (15.1) gives \(\dfrac{1}{\lambda} = \dfrac{R}{n^{2}}\), from which the result follows at once.
Notice that the series limit depends only on \(n\). Since \(n\) is larger for the later series, their limiting wavelengths are longer, which is why the Lyman series sits in the ultraviolet and the Paschen series in the infrared.
Every Element Has Its Own Set of Lines
Atoms of other elements also emit line spectra. The wavelengths of the lines emitted by each element are unique — so much so that we can identify the element from the wavelengths of the spectral lines that it emits. A spectrum therefore acts as a chemical label that can be read from a distance, without any sample being handled.
Rutherford’s model could not explain the atomic spectra. It offered no reason why an atom should be restricted to a handful of wavelengths, no reason why those wavelengths should fall into neat series, and certainly no reason why they should obey a formula as simple as Eq. (15.1). An empirical formula that works perfectly but has no explanation behind it is an invitation to a new model, and that is what comes next.
15.6 Bohr’s Atomic Model
Keeping What Worked, Repairing What Did Not
Niels Bohr modified Rutherford’s model by applying ideas of quantum physics which were being developed at that time. He did not throw the nuclear picture away. He realized that Rutherford’s model is essentially correct and that all it needs is stability of the orbits.
Two repairs were required. First, the electron must be allowed to occupy only certain orbits that do not shrink. Second, the electrons in these stable orbits should not emit electromagnetic waves, as would be required by classical (Maxwell’s) electromagnetic theory for any accelerating charge.
This is a bold step, and it is worth being honest about what it is. Bohr did not derive these conditions from classical physics; he imposed them, because they were what the observed facts demanded. He made three postulates which defined his atomic model. These are given below.
The First Postulate
The electrons revolve around the nucleus in circular orbits.
This is the same assumption as in Rutherford’s model, and the mechanics behind it is unchanged: the centripetal force necessary for the circular motion is provided by the electrostatic force of attraction between the electron and the nucleus. Nothing new is being claimed here — Bohr simply retains the nuclear atom as his starting point.
The Second Postulate
The radius of the orbit of an electron can only take certain fixed values, such that the angular momentum of the electron in these orbits is an integral multiple of \(\dfrac{h}{2\pi}\), \(h\) being Planck’s constant.
In symbols, the allowed angular momentum is \(\dfrac{nh}{2\pi}\), where \(n\) is a positive integer (a pure number) and \(h = 6.626 \times 10^{-34}\) J s. Angular momentum itself has the SI unit kg m\(^2\) s\(^{-1}\), which is the same as J s.
Such orbits are called stable orbits or stable states of the electrons, and electrons in these orbits do not emit radiation, as is demanded by classical physics. This is precisely the point at which the collapse predicted in the previous section is blocked, as indicated in Fig. 15.6a.
Diagram
Fig. 15.6a – A nucleus with a few allowed circular orbits drawn as solid circles labelled n = 1, 2, 3 and the forbidden in-between radii shown as faint dashed circles.
Thus, different orbits have different and definite values of angular momentum, and therefore different values of energies. This is the key structural idea: an electron in an atom cannot have just any energy, but only one of a definite set of values fixed by the integer \(n\).
The Third Postulate
An electron can make a transition from one of its orbits to another orbit having lower energy. In doing so, it emits a photon of energy equal to the difference in its energies in the two orbits.
A photon is a single quantum, or packet, of electromagnetic radiation, and its energy is related to its frequency \(\nu\) by \(E = h\nu\). So if the electron falls from an orbit of energy \(E_m\) to one of lower energy \(E_n\), the emitted photon carries energy \(h\nu = E_m - E_n\), with \(E\) in joule (J) and \(\nu\) in hertz (Hz).
Diagram
Fig. 15.6b – An electron jumping from an outer allowed orbit to an inner one, with a wavy arrow leaving the atom representing the emitted photon.
Notice how this postulate turns the earlier failure into a success. Radiation is no longer emitted continuously; it is emitted only during a jump. And since the orbit energies form a fixed set, the possible energy differences also form a fixed set, so only certain photon energies — that is, only certain frequencies and wavelengths — can ever come out of the atom.
The three postulates are frequently asked for as a direct question, and they are also the starting point of the derivations that follow, so state them carefully and in order — circular orbit and its force condition, quantisation of angular momentum with the non-radiating stable orbits, and emission of a photon during a downward transition.
15.6.1 Radii of the Orbits
Setting Up the Two Equations
Using the first two postulates we can study the entire dynamics of the circular motion of the electron, including its energy. The strategy is straightforward: write down one equation from mechanics and one from the quantum condition, and then solve them together.
Let the mass of the electron be \(m_e\) (in kg), its velocity in the \(n^{\text{th}}\) stable orbit be \(v_n\) (in m s\(^{-1}\)) and the radius of its orbit be \(r_n\) (in m). The angular momentum of the electron is then \(m_e v_n r_n\), and according to the second postulate above, we can write
\[m_e v_n r_n = \frac{nh}{2\pi} \qquad ---(15.2)\]
The positive integer \(n\) is called the principal quantum number of the electron.
The second equation comes from the first postulate. The centripetal force necessary for the circular motion of the electron is provided by the electrostatic force of attraction between the electron and the nucleus. Assuming the atomic number (number of electrons) of the atom to be \(Z\), the total positive charge on the nucleus is \(Ze\), and we can write
\[\frac{m_e v_n^{2}}{r_n} = \frac{Ze^{2}}{4\pi \varepsilon_0 r_n^{2}} \qquad ---(15.3)\]
Here \(\varepsilon_0\) is the permittivity of free space (vacuum), with the SI unit C\(^2\) N\(^{-1}\) m\(^{-2}\), and \(e\) is the electron charge in coulomb (C). The left-hand side is the centripetal force \(\dfrac{m_e v_n^{2}}{r_n}\) required to keep the electron on its circular path, and the right-hand side is the Coulomb force of attraction between the nuclear charge \(Ze\) and the electronic charge \(e\) separated by a distance \(r_n\).
Diagram
Fig. 15.6c – An electron of mass \(m_e\) moving with speed \(v_n\) along a circular orbit of radius \(r_n\) around a nucleus of charge \(+Ze\), with the electrostatic force marked pointing towards the nucleus and the velocity marked along the tangent.
Solving for the Radius
We now have two equations, (15.2) and (15.3), containing two unknowns, \(v_n\) and \(r_n\). Eliminating \(v_n\) between them gives the radius of the orbit, and eliminating \(r_n\) gives the speed of the electron.
From Eq. (15.2), \(v_n = \dfrac{nh}{2\pi m_e r_n}\). Substituting this into the left-hand side of Eq. (15.3) gives
\[\frac{m_e}{r_n}\left(\frac{nh}{2\pi m_e r_n}\right)^{2} = \frac{Ze^{2}}{4\pi \varepsilon_0 r_n^{2}}\]
that is,
\[\frac{n^{2}h^{2}}{4\pi^{2} m_e r_n^{3}} = \frac{Ze^{2}}{4\pi \varepsilon_0 r_n^{2}}\]
Cancelling \(r_n^{2}\) from both sides and rearranging for \(r_n\), we get
\[\therefore \; r_n = \frac{\varepsilon_0 n^{2} h^{2}}{\pi m_e Z e^{2}} \qquad ---(15.4)\]
Similarly, eliminating \(r_n\) from Eq. (15.2) and Eq. (15.3), we get
\[v_n = \frac{Ze^{2}}{2 \varepsilon_0 n h} \qquad ---(15.5)\]
Equation (15.5) is worth reading physically as well: the speed of the electron falls off as \(\dfrac{1}{n}\), so an electron in a distant orbit moves more slowly than one in a close orbit, much as a distant planet moves more slowly than a near one.
What Equation (15.4) Tells Us
Equation (15.4) shows that the radius of the orbit is proportional to \(n^{2}\), i.e., the square of the principal quantum number, since every other symbol on the right-hand side is a constant for a given atom.
The radius therefore increases with increase in \(n\), and it does so rapidly: the second orbit is four times the first, the third nine times the first, and so on. The allowed orbits are thus not evenly spaced but lie further and further apart, as shown in Fig. 15.6d.
Diagram
Fig. 15.6d – The first four Bohr orbits of hydrogen drawn to scale, with radii in the ratio 1 : 4 : 9 : 16, showing the widening gaps between successive orbits.
The hydrogen atom has only one electron, i.e., \(Z\) is 1. Substituting the values of the constants \(h\), \(\varepsilon_0\), \(m\) and \(e\) in Eq. (15.4), we get, for \(n = 1\),
\[r_1 = 0.053 \text{ nm}\]
This is called the Bohr radius and is denoted by
\[a_0 = \frac{\varepsilon_0 h^{2}}{\pi m_e e^{2}}\]
This is the radius of the smallest orbit of the electron in a hydrogen atom. It is the closest the electron can come to the nucleus in this model, and it therefore serves as a natural measure of the size of the atom itself.
From Eq. (15.4), we can write
\[r_n = n^{2} a_0 \qquad ---(15.6)\]
This compact form is the one used in most problems on hydrogen. Note carefully that it already assumes \(Z = 1\). For any other one-electron system the orbit is smaller by a factor of \(Z\), because Eq. (15.4) gives \(r_n \propto \dfrac{n^{2}}{Z}\).
Questions on this result are usually set as ratios rather than as full substitutions, because all the constants cancel. If you are asked how much larger the \(4^{\text{th}}\) orbit is than the \(2^{\text{nd}}\), or which one-electron system has the smallest first orbit, the entire answer lies in remembering that \(r_n \propto \dfrac{n^{2}}{Z}\) and in reading \(Z\) correctly from the species named in the question.
15.6.2 Energy of the Electrons
Adding the Two Kinds of Energy
An electron moving in an orbit possesses energy of two kinds. It is moving, so it has kinetic energy; and it is held by the attraction of the nucleus, so it has electrostatic potential energy. The total energy of an orbiting electron is the sum of its kinetic energy and its electrostatic potential energy. Thus,
\[E_n = K.E. + P.E.\]
\(E_n\) being the total energy of an electron in the \(n^{\text{th}}\) orbit, measured in joule (J) or, more conveniently for atoms, in electron volt (eV).
\[E_n = \frac{1}{2} m_e v_n^{2} - \frac{Ze^{2}}{4\pi \varepsilon_0 r_n}\]
The first term is the kinetic energy \(\dfrac{1}{2} m_e v_n^{2}\) of the electron. The second term is its potential energy in the field of the nucleus, and it carries a negative sign because the force between the nucleus and the electron is attractive; the zero of potential energy is taken at infinite separation, so bringing the electron closer lowers its energy below zero.
Using Eq. (15.3) and (15.4) this gives
\[E_n = -\frac{m_e Z^{2} e^{4}}{8 \varepsilon_0^{2} h^{2} n^{2}} \qquad ---(15.7)\]
It is worth seeing how the two terms combine. Equation (15.3) can be rewritten as \(m_e v_n^{2} = \dfrac{Ze^{2}}{4\pi \varepsilon_0 r_n}\), so the kinetic energy is \(\dfrac{Ze^{2}}{8\pi \varepsilon_0 r_n}\) — exactly half the magnitude of the potential energy, and opposite in sign to it. Adding them leaves \(E_n = -\dfrac{Ze^{2}}{8\pi \varepsilon_0 r_n}\), and substituting \(r_n\) from Eq. (15.4) into this gives Eq. (15.7).
Reading the Negative Sign
The negative value of the energy of the electron indicates that the electron is bound inside the atom, and that it has to be given energy so as to make the total energy zero, i.e., to make the electron free from the nucleus.
The energy increases (becomes less negative) with increase in \(n\), because \(E_n\) varies as \(\dfrac{1}{n^{2}}\). An electron in a wider orbit is therefore held less tightly.
Substituting the values of the constants \(m_e\), \(h\), \(e\) and \(\varepsilon_0\) in the above equation, we get
\[E_n = -13.6 \, \frac{Z^{2}}{n^{2}} \text{ eV} \qquad ---(15.8)\]
This is the working form of the result. Here \(Z\) is the atomic number and \(n\) the principal quantum number, both pure numbers, and the answer comes out directly in electron volt.
Ground State and Excited States
The first orbit (\(n = 1\)), which has minimum energy, is called the ground state of the atom. Orbits with higher values of \(n\), and therefore higher values of energy, are called the excited states of the atom. If the electron is in the \(n^{\text{th}}\) orbit, it is said to be in the \(n^{\text{th}}\) energy state.
For hydrogen atom (\(Z = 1\)) the energy of the electron in its ground state is \(-13.6\) eV, and the energies of the excited states increase as given by Eq. (15.8). The energy levels of hydrogen atom are shown in Fig. 15.4, where the energies of the levels are given in eV and the arrows mark transitions between them.
Diagram
Fig. 15.4 – Energy level diagram of the hydrogen atom: horizontal lines for n = 1, 2, 3, 4 … labelled −13.6 eV, −3.4 eV, −1.51 eV, −0.85 eV … converging towards 0 eV, with downward arrows marking transitions of the Lyman, Balmer and Paschen series (energy not to scale).
The energy levels come closer and closer as \(n\) increases, and their energy reaches a limiting value of zero as \(n\) goes to infinity. This crowding of the levels at the top is the direct counterpart of the crowding of the spectral lines near the series limit that was noted in the previous section.
Excitation Energy
An atom does not jump to a higher level on its own; it must be supplied with exactly the right amount of energy. The energy required to take an electron from the ground state to an excited state is called the excitation energy of the electron in that state.
For hydrogen atom, the minimum excitation energy (of the \(n = 2\) state) is \(-3.4 - (-13.6) = 10.2\) eV. Note how the subtraction works: both energies are negative, and the difference between them is positive, as it must be for energy that has to be supplied.
Ionization Energy
In order to remove or take out the electron in the ground state from a hydrogen atom, i.e., to make it free (and have zero energy), we have to supply 13.6 eV energy to it. This energy is called the ionization energy of the hydrogen atom.
It is the binding energy of the hydrogen atom. The two names describe the same quantity from opposite directions: energy that must be put in to pull the atom apart, and energy that was given out when it was assembled. If we form a hydrogen atom by bringing a proton and an electron from infinity and combining them, 13.6 eV energy will be released.
Be careful to keep excitation energy and ionization energy apart in problems. Excitation energy is a difference between two bound levels, whereas ionization energy is the difference between the ground state and the zero level at \(n = \infty\). Both are obtained from Eq. (15.8), but the second one always uses \(E_{\infty} = 0\).
Wavelengths of the Emitted Lines
According to the third postulate of Bohr, when an electron makes a transition from the \(m^{\text{th}}\) to the \(n^{\text{th}}\) orbit (\(m > n\)), the excess energy \(E_m - E_n\) is emitted in the form of a photon.
The energy of the photon, which can be written as \(h\nu\), \(\nu\) being its frequency, is therefore given by
\[h\nu = \frac{m_e Z^{2} e^{4}}{8 \varepsilon_0^{2} h^{2}} \left(\frac{1}{n^{2}} - \frac{1}{m^{2}}\right)\]
This follows by writing Eq. (15.7) once for \(m\) and once for \(n\) and subtracting; the two negative signs combine to leave the bracket in the order shown, which is positive because \(m > n\).
Using \(\nu = \dfrac{c}{\lambda}\), this can be written in terms of the wavelength as
\[\frac{1}{\lambda} = \frac{m_e Z^{2} e^{4}}{8 \varepsilon_0^{2} c h^{3}} \left(\frac{1}{n^{2}} - \frac{1}{m^{2}}\right) \qquad ---(15.9)\]
Here \(c\) is the velocity of light in vacuum, in m s\(^{-1}\).
We define a constant called the Rydberg’s constant, \(R_H\), as
\[R_H = \frac{m_e e^{4}}{8 \varepsilon_0^{2} c h^{3}} = 1.097 \times 10^{7} \text{ m}^{-1} \qquad ---(15.10)\]
In terms of \(R_H\), the wavelength is given by
\[\frac{1}{\lambda} = R_H Z^{2} \left(\frac{1}{n^{2}} - \frac{1}{m^{2}}\right) \qquad ---(15.11)\]
This is called the Rydberg’s formula. Remember that for hydrogen \(Z\) is 1.
Thus, Eq. (15.11) correctly describes the observed spectrum of hydrogen as given by Eq. (15.1). This is the moment at which Bohr’s model proves itself. The empirical relation of Section 15.5, which had been extracted from measurements with no theory behind it, now emerges as a consequence of the postulates — and even the numerical value of its constant is predicted from \(m_e\), \(e\), \(\varepsilon_0\), \(c\) and \(h\) alone.
15.6.3 Limitations of Bohr’s Model
A Model That Succeeded Only for the Simplest Atom
Even though Bohr’s model seemed to explain the hydrogen spectrum, it had a few shortcomings, which are listed below. Two of them are experimental — things the model gets wrong or cannot describe — and the third is a difficulty of principle.
It could not explain the line spectra of atoms other than hydrogen. Even for hydrogen, a more accurate study of the observed spectra showed multiple components in some lines, which could not be explained on the basis of this model.
The intensities of the emission lines seemed to differ from line to line, and Bohr’s model had no explanation for that.
On the theoretical side also the model was not entirely satisfactory, as it arbitrarily assumed orbits following a particular condition to be stable. There was no theoretical basis for that assumption.
Understanding Each Shortcoming
The first shortcoming is the most serious in practice. Bohr’s derivation assumed a single electron moving in the field of a nucleus of charge \(Ze\). In an atom with more than one electron, each electron is also repelled by the other electrons, and that repulsion depends on where those electrons happen to be. The simple force equation, Eq. (15.3), is then no longer valid, and the neat result \(E_n \propto \dfrac{Z^{2}}{n^{2}}\) does not carry over. This is why the model works for hydrogen and for one-electron ions such as He\(^{+}\), but fails for helium itself and for every heavier atom.
The mention of “multiple components” refers to what is seen when the hydrogen spectrum is examined with a high-resolution instrument. What appears as a single line under an ordinary spectrometer is then found to be two or more very closely spaced lines. Since Bohr’s model assigns exactly one energy to each value of \(n\), it can produce only one line for each transition and has no room for such splitting.
The third shortcoming is of a different character. Nothing in the model explains why the angular momentum should be restricted to integral multiples of \(\dfrac{h}{2\pi}\). The condition was chosen because it gives the right answers, not because it follows from any deeper principle — and a rule adopted only because it works is not yet physics.
Where the Explanation Actually Lies
A full quantum mechanical study is required for the complete understanding of the structure of atoms, which is beyond the scope of this book. In that treatment the electron is not a particle on a track at all, and the allowed energies emerge as a natural consequence rather than being assumed.
Some reasoning for the third shortcoming — that is, a theoretical basis for the second postulate in Bohr’s atomic model — was given by de Broglie, which we consider next.
When a question asks for the limitations of Bohr’s model, give all three points and keep them distinct: failure for multi-electron atoms together with the unexplained splitting of lines, the unexplained differences in intensity, and the arbitrary nature of the quantisation condition itself.
15.6.4 De Broglie’s Explanation
From Particle on a Track to Wave on a Loop
We have seen in Chapter 14 that material particles also have a dual nature, like that of light, and that there is a wave associated with every material particle. An electron, therefore, need not always be pictured as a tiny ball; in some situations it behaves as a wave, with a wavelength of its own.
De Broglie suggested that instead of considering the orbiting electrons inside atoms as particles, we should view them as standing waves. This single change of picture is what supplies the missing justification for Bohr’s second postulate.
Recall what a standing wave requires. Similar to the case of standing waves on strings or in pipes, as studied in Chapter 6, a standing wave can exist only when the wave fits exactly into the space available to it. For an electron the “space available” is its own orbit, which closes on itself, so the length of the orbit of an electron has to be an integral multiple of its wavelength.
Thus, the length of the first orbit will be equal to one de Broglie wavelength \(\lambda_1\) of the electron in that orbit, that of the second orbit will be twice the de Broglie wavelength of the electron in that orbit, and so on. This is shown for the 4th orbit in Fig. 15.5, where exactly four wavelengths fit around the circle.
Diagram
Fig. 15.5 – Standing electron wave for the 4th orbit of an electron in an atom: a circular orbit around the nucleus with exactly four complete wavelengths fitted around its circumference, the wave joining smoothly onto itself.
Diagram
Fig. 15.6e – Comparison of two circular orbits: one containing a whole number of wavelengths with the wave joining smoothly onto itself, and one containing a non-integral number where the wave fails to match up and cancels.
Turning the Picture into Bohr’s Condition
In general, we can write
\[2\pi r_n = n \lambda_n, \qquad n = 1, 2, 3, \ldots\]
giving
\[\lambda_n = \frac{2\pi r_n}{n} \qquad ---(15.12)\]
Here \(2\pi r_n\) is the circumference of the \(n^{\text{th}}\) orbit in metre (m), \(\lambda_n\) is the de Broglie wavelength of the electron in that orbit in metre (m), and \(n\) is a positive integer.
The de Broglie wavelength is related to the linear momentum \(p_n\) of the particle by
\[\lambda_n = \frac{h}{p_n} = \frac{h}{m v_n}\]
with \(p_n\) in kg m s\(^{-1}\) and \(h\) in J s. This relation was introduced in Chapter 14 and is what connects the wave description to the particle description of the same electron.
Substituting this in Eq. (15.12) gives
\[p_n = \frac{nh}{2\pi r_n}\]
Thus, the angular momentum of the electron in the \(n^{\text{th}}\) orbit, \(L_n\), can be written as
\[L_n = p_n r_n = \frac{nh}{2\pi}\]
which is the second postulate of Bohr’s atomic model.
Therefore, considering electrons as waves gives some theoretical basis for the second postulate made by Bohr. What Bohr had to assume — that angular momentum comes only in multiples of \(\dfrac{h}{2\pi}\) — now appears as a consequence of a more general idea about matter, namely that every particle has a wave associated with it. Note the careful wording: this is a reason for the postulate, not a full derivation of atomic structure, which still requires the complete quantum mechanical treatment.
15.7 Atomic Nucleus
Turning Attention Inward
Everything discussed so far has been about the electrons. The nucleus entered the story only as a small, heavy, positively charged object placed at the centre — a source of the attractive force, and nothing more. That was enough to explain the size of the atom, its stability and its spectrum.
It is not enough to explain the atom itself. The nucleus carries almost the whole mass of the atom, it fixes the atomic number and hence the chemistry of the element, and it is the seat of processes that release energies a million times larger than any chemical reaction can. From this section onwards, therefore, the nucleus is treated not as a point charge but as an object with its own structure, its own size, its own internal forces and its own ways of changing.
Four questions organise the sections that follow. What is a nucleus made of? How large is it? What holds it together, given that its positive charges repel one another fiercely? And how much energy is locked in that binding? Once these are answered, radioactive decay and nuclear energy follow naturally, because both are simply the nucleus rearranging itself towards a more tightly bound state.
Diagram
Fig. 15.7 – An atom shown at two scales: the full atom with its electron orbits, and beside it a magnified view of the central nucleus showing protons and neutrons packed closely together.
Keep one contrast in mind throughout this half of the chapter. Atomic processes involving electrons deal in energies of a few electron volts; nuclear processes deal in energies of millions of electron volts. Nearly every distinctive feature of nuclear physics — the penetrating radiation, the enormous energy release, the extreme density — traces back to that factor of a million.
15.7.1 Constituents of a Nucleus
The Two Particles Inside
The atomic nucleus is made up of subatomic — meaning smaller than an atom — particles called protons and neutrons. Together, protons and neutrons are referred to as nucleons.
The mass of a proton is about 1836 times that of an electron. This is why almost the entire mass of an atom belongs to its nucleus, even though the nucleus occupies a negligible fraction of the atom’s volume. The mass of a neutron is nearly the same as that of a proton, but is slightly higher.
The proton is a positively charged particle. The magnitude of its charge is equal to the magnitude of the charge of an electron, so a proton and an electron neutralise each other exactly. The neutron, as the name suggests, is electrically neutral.
Diagram
Fig. 15.7a – A nucleus drawn as a cluster of closely packed protons (shaded, marked +) and neutrons (unshaded), with labels showing Z protons, N neutrons and total A = Z + N nucleons.
Counting the Nucleons: Z, N and A
The number of protons in an atom is called its atomic number and is designated by \(Z\). The number of electrons in a neutral atom is also equal to \(Z\). Thus, the total positive and total negative charges in an atom are equal in magnitude, and the atom as a whole is electrically neutral.
The number of neutrons in a nucleus is written as \(N\). The total number of nucleons in a nucleus is called the mass number of the atom and is designated by
\[A = Z + N\]
The mass number determines the mass of a nucleus and of the atom. Since every nucleon has a mass close to one unified atomic mass unit, an atom of mass number \(A\) has a mass of roughly \(A\) such units.
The atoms of an element X are represented by the symbol for the element together with its atomic and mass numbers, written as \(^{A}_{Z}\text{X}\). The mass number is written above and the atomic number below, both on the left of the symbol. For example, the symbols for hydrogen, carbon and oxygen atoms are written as \(^{1}_{1}\text{H}\), \(^{12}_{6}\text{C}\) and \(^{16}_{8}\text{O}\).
The chemical properties of an atom are decided by the number of electrons present in it, i.e., by \(Z\). Chemistry is the behaviour of the outer electrons; the nucleus only decides how many electrons there will be.
Isotopes, Isobars and Isotones
The number of protons and electrons in the atoms of a given element are fixed. For example, a hydrogen atom has one proton and one electron; a carbon atom has six protons and six electrons.
The number of neutrons in the atoms of a given element, however, can vary. For example, a hydrogen nucleus can have zero, one or two neutrons. These varieties of hydrogen are referred to as \(^{1}_{1}\text{H}\), \(^{2}_{1}\text{H}\) and \(^{3}_{1}\text{H}\) and are respectively called hydrogen, deuterium and tritium, as shown in Fig. 15.7b.
Diagram
Fig. 15.7b – The three varieties of hydrogen: a nucleus with one proton, one with a proton and a neutron, and one with a proton and two neutrons, labelled hydrogen, deuterium and tritium, each with a single electron around it.
Atoms having the same number of protons but a different number of neutrons are called isotopes. Thus, deuterium and tritium are isotopes of hydrogen. They have the same chemical properties as those of hydrogen.
Similarly, a helium nucleus can have one or two neutrons, and these are referred to as \(^{3}_{2}\text{He}\) and \(^{4}_{2}\text{He}\).
The atoms having the same mass number \(A\) are called isobars. Thus, \(^{3}_{1}\text{H}\) and \(^{3}_{2}\text{He}\) are isobars.
Atoms having the same number of neutrons but different values of atomic number \(Z\) are called isotones. Thus, \(^{3}_{1}\text{H}\) and \(^{4}_{2}\text{He}\) are isotones.
A quick way to keep the three apart: isotopes share \(Z\), isobars share \(A\), and isotones share \(N\). Check the last pair above — \(^{3}_{1}\text{H}\) has \(N = 3 - 1 = 2\) and \(^{4}_{2}\text{He}\) has \(N = 4 - 2 = 2\), so they are indeed isotones.
Units for Measuring Masses of Atoms and Subatomic Particles
Masses of atoms and subatomic particles are measured in three different units.
The first unit is the usual unit kg. The masses of the electron, proton and neutron, \(m_e\), \(m_p\) and \(m_n\) respectively, in this unit are:
\[m_e = 9.109383 \times 10^{-31} \text{ kg}\] \[m_p = 1.672623 \times 10^{-27} \text{ kg}\] \[m_n = 1.674927 \times 10^{-27} \text{ kg}\]
Obviously, kg is not a convenient unit to measure the masses of atoms or subatomic particles, which are extremely small compared to one kg. Working with such numbers means carrying awkward powers of ten through every calculation, and small differences between them — which is exactly what matters in nuclear physics — become hard to handle.
Therefore, another unit called the unified atomic mass unit (u) is used for the purpose. One u is equal to \(\dfrac{1}{12}\)th of the mass of a neutral carbon atom of mass number 12, in its lowest electronic state, and
\[1 \text{ u} = 1.6605402 \times 10^{-27} \text{ kg}\]
In this unit, the masses of the above three particles are
\[m_e = 0.00055 \text{ u}\] \[m_p = 1.007825 \text{ u}\] \[m_n = 1.008665 \text{ u}\]
Notice how convenient this is: each nucleon comes out close to 1 u, so the mass of an atom in u is close to its mass number \(A\). The small departures from whole numbers are not rounding errors — they are physically important, and the next section will show that they are the key to nuclear binding energy.
The third unit for measuring masses of atoms and subatomic particles is in terms of the amount of energy that their masses are equivalent to. According to Einstein’s famous mass-energy relation, a particle having mass \(m\) is equivalent to an amount of energy
\[E = mc^{2}\]
where \(c\) is the velocity of light in vacuum in m s\(^{-1}\), \(m\) is in kg and \(E\) is in joule (J).
The unit used to measure masses in terms of their energy equivalent is the eV/\(c^2\). Writing the mass as an energy divided by \(c^{2}\) keeps the relation \(E = mc^2\) ready for immediate use, which is why nuclear physicists prefer it. One atomic mass unit is equal to 931.5 MeV/\(c^2\).
The masses of the three particles in this unit are
\[m_e = 0.511 \text{ MeV}/c^{2}\] \[m_p = 938.28 \text{ MeV}/c^{2}\] \[m_n = 939.57 \text{ MeV}/c^{2}\]
The conversion factor 931.5 MeV per u is one of the most frequently used numbers in this chapter. Whenever a mass difference in u has been worked out, multiplying it by 931.5 gives the corresponding energy directly in MeV, and almost every numerical problem on binding energy, decay energy and nuclear energy ends with that step.
15.7.2 Sizes of Nuclei
What Decides the Size of a Nucleus
The size of an atom is decided by the sizes of the orbits of the electrons in the atom. The larger the number of electrons in an atom, the higher are the orbits occupied by them, and the larger is the size of the atom.
Similarly, all nuclei do not have the same size. Obviously, the size of a nucleus depends on the number of nucleons present in it, i.e., on its mass number \(A\). A nucleus with more nucleons to accommodate must occupy more room.
From experimental observations it has been found that the radius \(R_X\) of a nucleus X is related to \(A\) as
\[R_X = R_0 A^{1/3} \qquad ---(15.13)\]
where \(R_0 = 1.2 \times 10^{-15}\) m. Here \(R_X\) is in metre (m) and \(A\) is a pure number, so \(R_0\) carries the unit of length; it is the radius a nucleus would have if it contained a single nucleon.
The cube root is the significant feature of this result. Since volume is proportional to the cube of the radius, \(R \propto A^{1/3}\) means that the volume of a nucleus is directly proportional to the number of nucleons in it. In other words, each nucleon occupies a fixed share of space, whatever nucleus it happens to be in, as suggested by the straight-line plot in Fig. 15.7c.
Diagram
Fig. 15.7c – Graph of nuclear radius R plotted against \(A^{1/3}\) for several nuclei, the points lying on a straight line through the origin whose slope is \(R_0\).
Density of Nuclear Matter
We can now calculate how tightly matter is packed inside a nucleus. Treating the nucleus as a sphere of radius \(R_X\), its volume is \(\dfrac{4}{3}\pi R_X^{3}\), and its mass is \(mA\), where we have assumed \(m\) to be the average mass of a nucleon (proton and neutron), as the difference in their masses is rather small.
The density is then given by
\[\rho = \frac{3mA}{4\pi R_X^{3}}\]
Substituting for \(R_X\) from Eq. (15.13), we get
\[\rho = \frac{3mA}{4\pi \left(R_0 A^{1/3}\right)^{3}} = \frac{3m}{4\pi R_0^{3}} = \text{constant}\]
The mass number cancels completely, because \(\left(A^{1/3}\right)^{3} = A\) in the denominator cancels the \(A\) in the numerator. Nothing containing \(A\) survives.
Thus, the density of a nucleus does not depend on the atomic number of the nucleus and is the same for all nuclei. A gold nucleus, a carbon nucleus and a uranium nucleus are all made of matter of exactly the same density; they differ only in how much of it there is.
Substituting the values of the constants \(m\), \(\pi\) and \(R_0\), the value of the density is obtained as \(2.3 \times 10^{17}\) kg m\(^{-3}\), which is extremely large.
To appreciate what this figure means, compare it with ordinary matter. Among all known elements, osmium is known to have the highest density, which is only \(2.2 \times 10^{4}\) kg m\(^{-3}\). This is smaller than the nuclear density by thirteen orders of magnitude — that is, by a factor of about \(10^{13}\).
The reason is the one already met in Section 15.4: an ordinary solid is mostly empty space, with tiny nuclei separated by the vast regions occupied by electrons. Nuclear matter is what remains when that emptiness is removed.
Note carefully, for objective questions, which nuclear properties depend on \(A\) and which do not: the radius, the volume and the mass all grow with \(A\), but the density is the same for every nucleus.
15.7.3 Nuclear Forces
A Fourth Force, Acting Only Inside the Nucleus
You have learnt about the four fundamental forces that occur in nature. Out of these four, the force that determines the structure of the nucleus is the strong force, also called the nuclear force. This acts between protons and neutrons and is mostly attractive.
It is different from the electrostatic and gravitational force in terms of its strength and range, i.e., the distance up to which it is effective. Over short distances of about a few fm, the strength of the nuclear force is much higher than that of the other two forces.
Its range, however, is very small, and its strength goes to zero when two nucleons are at a distance larger than a few fm. This is in contrast to the ranges of electrostatic and gravitational forces, which are infinite. Those two forces weaken gradually as the inverse square of the distance, but they never quite vanish; the nuclear force simply switches off beyond a few femtometres, as shown in Fig. 15.7d.
Diagram
Fig. 15.7d – Graph of the nuclear force between two nucleons against their separation in fm, showing a strong attraction over a few fm that falls rapidly to zero at larger separations, compared with the slowly decreasing electrostatic force.
How the Nucleus Survives Its Own Repulsion
The protons in the nucleus repel one another due to their similar (positive) charges. At separations of a few femtometres this repulsion is enormous, because the Coulomb force varies as the inverse square of the distance and these distances are extremely small.
The nuclear forces between the nucleons counter the forces of electrostatic repulsion. As the nuclear force is much stronger than the electrostatic force for the distances between nucleons in a typical nucleus, it overcomes the repulsive force and keeps the nucleons together, making the nucleus stable.
Notice also the role played by the neutrons. They carry no charge, so they add attraction without adding any repulsion — which is why heavier nuclei, with more protons pushing each other apart, need proportionately more neutrons to stay bound.
Properties of the Nuclear Force
The nuclear force is not yet well understood. Unlike the electrostatic force, it cannot be written down as one simple formula that holds at all separations. What we know about its properties can be summarized as follows.
It is the strongest force among subatomic particles. Its strength is about 50–60 times larger than that of the electrostatic force.
Unlike the electromagnetic and gravitational forces, which act over large distances (their range is infinity), the nuclear force has a range of about a few fm, and the force is negligible when two nucleons are separated by larger distances.
The nuclear force is independent of the charge of the nucleons, i.e., the nuclear force between two neutrons with a given separation is the same as that between two protons, or between a neutron and a proton, at the same separation.
The third property deserves a moment’s thought. It does not say that the total force between two protons equals that between two neutrons — the two protons also repel each other electrostatically. It says that the nuclear part of the interaction does not care about charge at all. As far as the strong force is concerned, a proton and a neutron are the same kind of object.
These three properties — greatest strength, very short range, and charge independence — are the standard answer to a direct question on the nuclear force, so learn them as a set and be ready to contrast each one with the corresponding property of the electrostatic force.
15.8 Nuclear Binding Energy
From the Atom to the Nucleus
We have seen that in a hydrogen atom, the energy with which the electron in its ground state is bound to the nucleus (which is a single proton in this case) is 13.6 eV. This is the amount of energy which is released when a proton and an electron are brought from infinity to form the atom in its ground state. In other words, this is the amount of energy which has to be supplied to the atom to separate the electron and the proton, i.e., to make them free.
The same idea can be applied one level deeper. The protons and the neutrons inside a nucleus are also bound to one another, held by the nuclear force described in the previous section. Energy has to be supplied to the nucleus to make the nucleons free, i.e., to separate them and take them to large distances from one another. This energy is the binding energy of the nucleus.
The same amount of energy is released if we bring individual nucleons from infinity to form the nucleus. Binding energy is therefore not something stored “in” the nucleus like a fuel; it is the energy debt that must be paid to take the nucleus apart, and equally the energy that was given out when it was assembled.
Where the Energy Comes From: Mass Defect
Where does this released energy come from? It comes from the masses of the nucleons.
The mass of a nucleus is smaller than the total mass of its constituent nucleons. Let the mass of a nucleus having atomic number \(Z\) and mass number \(A\) be \(M\). It is smaller than the sum of the masses of \(Z\) protons and \(N \; (= A - Z)\) neutrons. We can write,
\[\Delta M = Z m_p + N m_n - M \qquad ---(15.14)\]
\(\Delta M\) is called the mass defect of the nucleus. All the masses here may be taken in u, or in kg, as long as the same unit is used throughout.
The binding energy \(E_B\) of the nucleus is given by
\[E_B = \Delta M c^{2} = \left(Z m_p + N m_n - M\right)c^{2} \qquad ---(15.15)\]
This follows directly from Einstein’s mass-energy relation: the missing mass has not disappeared, it has been released as energy during the formation of the nucleus.
Using Atomic Masses Instead of Nuclear Masses
There is a practical difficulty with Eq. (15.15). What is tabulated and measured with high accuracy is the mass of the neutral atom, which includes its electrons, not the mass of the bare nucleus.
On the right hand side of Eq. (15.15), we can add and subtract the mass of \(Z\) electrons, which will enable us to use atomic masses in the calculation of binding energy. Equation (15.15) thus becomes
\[E_B = \left[\left(Z m_p + Z m_e\right) + N m_n - \left(M + Z m_e\right)\right]c^{2}\]
\[E_B = \left[Z m_H + N m_n - {}^{A}_{Z}M\right]c^{2} \qquad ---(15.16)\]
Here \(m_H\) is the mass of a hydrogen atom, and \({}^{A}_{Z}M\) is the atomic mass of the element being considered. The grouping works because one proton together with one electron is exactly a hydrogen atom, and the nucleus together with its \(Z\) electrons is exactly the neutral atom.
We will be using atomic masses in what follows, unless otherwise specified.
Binding Energy per Nucleon
An important quantity in this regard is the binding energy per nucleon \(\left(= \dfrac{E_B}{A}\right)\) of a nucleus. This can be considered to be the average energy which has to be supplied to a nucleon to remove it from the nucleus and make it free.
This quantity thus allows us to compare the relative strengths with which nucleons are bound in a nucleus for different species, and therefore to compare their stabilities. Nuclei with higher values of \(\dfrac{E_B}{A}\) are more stable as compared to nuclei having smaller values of this quantity.
The reason for dividing by \(A\) is worth stating. A heavy nucleus naturally has a larger total binding energy simply because it has more nucleons, so total binding energy is a poor guide to stability. Dividing by the number of nucleons removes this effect and gives a fair comparison between a light nucleus and a heavy one.
Binding energy per nucleon for different values of \(A\) (i.e., for nuclei of different elements) are plotted in Fig. 15.6.
Diagram
Fig. 15.6 – Graph of binding energy per nucleon (in MeV) against mass number A, rising steeply from deuterium, peaking near A = 56 at about 8.8 MeV, and falling slowly beyond A = 80.
Reading the Curve
The shape of this curve carries a great deal of information, and it is worth going along it carefully.
The deuterium nucleus has the minimum value of \(\dfrac{E_B}{A}\) and is therefore the least stable nucleus. With only two nucleons, each one has just a single partner to be attracted by.
The value of \(\dfrac{E_B}{A}\) increases with increase in atomic number and reaches a plateau for \(A\) between 50 to 80. Thus, the nuclei of these elements are the most stable among all the species.
The peak occurs around \(A = 56\), corresponding to iron, which is thus one of the most stable nuclei. To the right of the peak the curve turns over: the value of \(\dfrac{E_B}{A}\) decreases gradually for values of \(A\) greater than 80, making the nuclei of those elements slightly less stable. The reason is the one anticipated in the previous section — in a large nucleus the short-range nuclear attraction reaches only to a nucleon’s near neighbours, while the electrostatic repulsion acts between every pair of protons in the whole nucleus.
Note that the binding energy of the hydrogen nucleus, having a single proton, is zero. There is nothing to bind: with only one nucleon there is no second particle to be held to, so no energy is needed to “separate” it.
The fall in \(\dfrac{E_B}{A}\) on both sides of the peak is the single most useful feature of this graph, and later sections of this chapter return to it repeatedly, since it is what makes energy release from nuclear reactions possible at all.
15.9 Radioactive Decays
Why Some Nuclei Change on Their Own
Many of the nuclei are stable, i.e., they can remain unchanged for a very long time. These have a particular ratio of the mass number and the atomic number — that is, a particular balance between neutrons and protons that allows the nuclear attraction to hold the electrostatic repulsion in check.
Other nuclei occurring in nature are not so stable and undergo changes in their structure by emission of some particles. They change or decay to other nuclei (with different \(A\) and \(Z\)) in the process.
The decaying nucleus is called the parent nucleus, while the nucleus produced after the decay is called the daughter nucleus. The process is called radioactive decay or radioactivity, and was discovered by Becquerel (1852–1908) in 1896.
Radioactive decays occur because the parent nuclei are unstable and get converted to more stable daughter nuclei by the emission of some particles. The driving idea is the one met in the previous section: the system moves towards an arrangement in which its nucleons are more tightly bound, and the surplus energy is carried away by whatever is emitted.
These decays are of three types, as described below.
Alpha Decay
In this type of decay, the parent nucleus emits an alpha particle, which is the nucleus of a helium atom. The parent nucleus thus loses two protons and two neutrons.
The decay can be expressed as
\[{}^{A}_{Z}\text{X} \rightarrow {}^{A-4}_{Z-2}\text{Y} + \alpha \qquad ---(15.17)\]
X is the parent nucleus and Y is the daughter nucleus. Note how the bookkeeping works: the mass number falls by 4 because four nucleons leave together, and the atomic number falls by 2 because two of those nucleons are protons. The daughter is therefore an atom of a different element, two places earlier in the periodic table.
All nuclei with \(A > 210\) undergo alpha decay. The reason is that these nuclei have a large number of protons. The electrostatic repulsion between them is very large, and the attractive nuclear forces between the nucleons are not able to cope with it.
This is the short range of the nuclear force showing itself once again. In a very large nucleus, a proton is attracted only by the nucleons in its immediate neighbourhood, but it is repelled by every other proton in the nucleus. Beyond a certain size, repulsion wins. This makes the nucleus unstable, and it tries to reduce the number of its protons by ejecting them in the form of alpha particles.
An example of this is the alpha decay of bismuth, which is the parent nucleus with \(A = 212\) and \(Z = 83\). The daughter nucleus has \(A = 208\) and \(Z = 81\), which is thallium. The reaction is
\[{}^{212}_{83}\text{Bi} \rightarrow {}^{208}_{81}\text{T}\ell + \alpha\]
Diagram
Fig. 15.9a – Alpha decay represented as a heavy parent nucleus emitting a cluster of two protons and two neutrons, with the daughter nucleus labelled A − 4 and Z − 2.
Energy Released in Alpha Decay
The total mass of the products of an alpha decay is always less than the mass of the parent atom. The excess mass appears as the kinetic energy of the products — chiefly of the alpha particle, which is much the lighter of the two and therefore carries away most of the speed.
The difference in the energy equivalent of the mass of the parent atom and that of the sum of masses of the products is called the \(Q\)-value, \(Q\), of the decay, and is equal to the kinetic energy of the products. We can write,
\[Q = \left[m_X - m_Y - m_{He}\right]c^{2} \qquad ---(15.18)\]
\(m_X\), \(m_Y\) and \(m_{He}\) being the masses of the parent atom, the daughter atom and the helium atom. Note that we have used atomic masses to calculate the \(Q\) factor. This is legitimate because the electrons balance on the two sides: the parent atom has \(Z\) electrons, while the daughter atom has \(Z - 2\) and the helium atom has 2.
A positive \(Q\) is exactly the condition for the decay to be possible at all. If the products were heavier than the parent, energy would have to be supplied from outside and the decay could not occur spontaneously.
Beta Decay
In this type of decay the nucleus emits an electron, produced by converting a neutron in the nucleus into a proton. Thus, the basic process which takes place inside the parent nucleus is
\[n \rightarrow p + e^{-} + \text{antineutrino} \qquad ---(15.19)\]
It is important to be clear about where this electron comes from. It is not one of the orbital electrons of the atom, and it was not sitting inside the nucleus beforehand. It is created at the moment a neutron turns into a proton.
Neutrino and antineutrino are particles which have very little mass and no charge. They interact with matter extremely weakly, which is why they escape from the decay almost unnoticed, carrying away a share of the energy.
During beta decay, the number of nucleons, i.e., the mass number of the nucleus, remains unchanged, because one nucleon has merely changed its identity. The daughter nucleus has one less neutron and one extra proton. Thus, \(Z\) increases by one and \(N\) decreases by one, \(A\) remaining constant. The decay can be written as,
\[{}^{A}_{Z}\text{X} \rightarrow {}^{A}_{Z+1}\text{Y} + e^{-} + \text{antineutrino} \qquad ---(15.20)\]
An example is
\[{}^{60}_{27}\text{Co} \rightarrow {}^{60}_{28}\text{Ni} + e^{-} + \text{antineutrino}\]
Beta Plus Decay
There is another type of beta decay called the beta plus decay, in which a proton gets converted to a neutron by emitting a positron and a neutrino.
A positron is a particle with the same properties as an electron, except that its charge is positive. It is known as the antiparticle of the electron. This decay can be written as,
\[p \rightarrow n + e^{+} + \text{neutrino} \qquad ---(15.21)\]
The mass number remains unchanged during the decay, but \(Z\) decreases by one and \(N\) increases by one. The decay can be written as
\[{}^{A}_{Z}\text{X} \rightarrow {}^{A}_{Z-1}\text{Y} + e^{+} + \text{neutrino} \qquad ---(15.22)\]
An example is
\[{}^{22}_{11}\text{Na} \rightarrow {}^{22}_{10}\text{Ne} + e^{+} + \text{neutrino}\]
An interesting thing about beta plus decay is that the mass of a neutron is higher than the mass of a proton. Thus the decay described by Eq. (15.21) cannot take place for a free proton, since that would mean creating a heavier particle out of a lighter one with no source of the extra energy.
However, it can take place when the proton is inside a nucleus, as the extra energy needed to produce a neutron can be obtained from the rest of the nucleus. The nucleus as a whole ends up more tightly bound, and the energy released in that rearrangement pays for the conversion.
Energy Released in Beta Decay
In beta decay also, the total mass of the products of the decay is less than the mass of the parent atom. The excess mass is converted into the kinetic energy of the products. The \(Q\) value for the decay can be written as
\[Q = \left[m_X - m_Y - m_e\right]c^{2} \qquad ---(15.23)\]
Here, we have ignored the mass of the neutrino, as it is negligible compared to the masses of the nuclei.
Because the energy released is shared between the beta particle and the neutrino in varying proportions, the beta particle does not always come out with the same energy. The \(Q\) value gives the maximum energy the beta particle can have, which occurs when the neutrino happens to carry away nothing.
Gamma Decay
In this type of decay, gamma rays are emitted by the parent nucleus. As you know, a gamma ray is a high energy photon.
The daughter nucleus is the same as the parent nucleus, as no other particle is emitted, but it has less energy, since some energy goes out in the form of the emitted gamma ray. Neither \(A\) nor \(Z\) changes in gamma decay; only the internal energy state of the nucleus changes.
We have seen that the electrons in an atom are arranged in different energy levels (orbits), and that an electron from a higher orbit can make a transition to a lower orbit, emitting a photon in the process. The situation in a nucleus is similar. The nucleons occupy energy levels with different energies. A nucleon can make a transition from a higher energy level to a lower energy level, emitting a photon in the process.
Diagram
Fig. 15.9b – Nuclear energy level diagram showing an excited nucleus dropping to its ground state, either by a single gamma transition or by two successive transitions through an intermediate level.
The difference between atomic and nuclear energy levels is in their energies and energy separations. Energies and the differences in the energies of different levels in an atom are of the order of a few eVs, while those in the case of a nucleus are of the order of a few keV to a few MeV.
Therefore, whereas the radiations emitted by atoms are in the ultraviolet to radio region, the radiations emitted by nuclei are in the range of gamma rays. This single comparison — eV against MeV — is the reason nuclear radiation is so much more penetrating than anything an atom can emit.
When Gamma Decay Occurs
Usually, the nucleons in a nucleus are in the lowest possible energy state. They cannot easily get excited, as a large amount of energy (in keVs or MeVs) is required for their excitation, and ordinary processes such as heating cannot supply anything like that much.
A nucleon however may end up in an excited state as a result of the parent nucleus undergoing alpha or beta decay. Thus, gamma decays usually occur after one of these decays — the nucleus is left “shaken up” by the first decay and then settles down by emitting a gamma ray.
For example, \({}^{57}\text{Co}\) undergoes beta plus decay to form the daughter nucleus \({}^{57}\text{Fe}\), which is in an excited state having energy of 136 keV. There are two ways in which it can make a transition to its ground state. One is by emitting a gamma ray of energy 136 keV, and the other is by emitting a gamma ray of energy 122 keV and going to an intermediate state first, and then emitting a photon of energy 14 keV to reach the ground state. Both these emissions have been observed experimentally. Notice that the two routes add up to the same total, since 122 + 14 = 136 keV.
Decay Chains
Which type of decay a nucleus will undergo depends on which of the resulting daughter nuclei is more stable. The nucleus takes whichever route leads to a more tightly bound arrangement.
Often, the daughter nucleus is also not stable and it undergoes further decay. A chain of decays may take place until the final daughter nucleus is stable.
An example of such a series decay is that of \({}^{238}\text{U}\), which undergoes a series of alpha and beta decays, a total of 14 times, to finally reach a stable daughter nucleus of \({}^{206}\text{Pb}\).
Diagram
Fig. 15.9c – Schematic of the uranium-238 decay series, showing successive alpha and beta decays leading step by step to stable lead-206.
In numerical work on this section, the same routine applies to all three decays: write the balanced decay equation first, checking that \(A\) and \(Z\) each balance on the two sides, then find the mass difference in u, and finally multiply by 931.5 to get the energy released in MeV.
15.10 Law of Radioactive Decay
A Process That Cannot Be Predicted for a Single Atom
Materials which undergo alpha, beta or gamma decays are called radioactive materials.
The nature of radioactivity is such that if we have one atom of the radioactive material, we can never predict how long it will take to decay. It may decay in the next second, or it may survive for a thousand years. Nothing about the atom — its age, its history, its surroundings — makes any difference to the answer.
If we have \(N_0\) number of radioactive atoms (parent atoms or nuclei) of a particular kind, say uranium, at time \(t = 0\), all we can say is that their number will decrease with time as some nuclei — and we cannot say which ones — will decay.
Setting Up the Decay Law
Let us assume that at time \(t\), the number of parent nuclei which are left is \(N(t)\). How many of these will decay in the interval between \(t\) and \(t + dt\)?
We can guess that the larger the value of \(N(t)\), the larger will be the number of decays \(dN\) in time \(dt\) — twice as many nuclei present should give twice as many decays. Thus, we can say that \(dN\) will be proportional to \(N(t)\).
Also, we can guess that the larger the interval \(dt\), the larger will be the number of particles decaying in that interval. Thus, we can write,
\[dN \propto -N(t)\,dt\]
or,
\[dN = -\lambda N(t)\,dt \qquad ---(15.24)\]
where \(\lambda\) is a constant of proportionality called the decay constant. Its SI unit is s\(^{-1}\), since \(\lambda \, dt\) must be a pure number.
The negative sign in Eq. (15.24) indicates that the change in the number of parent nuclei \(dN\) is negative, i.e., \(N(t)\) is decreasing with time.
Integrating the Equation
Equation (15.24) tells us the rate of change of \(N\); to find \(N\) itself at any time we must integrate. Separating the variables, we can integrate this equation as
\[\int_{N_0}^{N(t)} \frac{dN}{N(t)} = -\lambda \int_{0}^{t} dt\]
Here, \(N_0\) is the number of parent atoms at time \(t = 0\); the limits on the two integrals correspond to each other, the number being \(N_0\) at time zero and \(N(t)\) at time \(t\).
Integration gives
\[\ln \frac{N(t)}{N_0} = -\lambda t\]
or,
\[N(t) = N_0 e^{-\lambda t} \qquad ---(15.25)\]
This is the decay law of radioactivity.
The exponential form is worth reading physically. The number does not fall by a fixed amount in each second; it falls by a fixed fraction. As the sample dwindles, the number of decays per second dwindles with it, so the curve of Fig. 15.10a approaches the time axis ever more gradually and, strictly speaking, never reaches it.
Diagram
Fig. 15.10a – Graph of the number of undecayed nuclei N(t) against time t, showing a smooth exponential fall from N₀ that flattens as it approaches the time axis.
Activity of a Sample
In an experiment we do not count the nuclei that remain; we count the decays as they happen, using a detector. This leads to a second quantity.
The rate of decay, i.e., the number of decays per unit time \(-\dfrac{dN(t)}{dt}\), also called the activity \(A(t)\), can be written using Eq. (15.24) and (15.25) as
\[A(t) = -\frac{dN}{dt} = \lambda N(t) = \lambda N_0 e^{-\lambda t} \qquad ---(15.26)\]
At \(t = 0\), the activity is given by \(A_0 = \lambda N_0\). Using this, Eq. (15.26) can be written as
\[A(t) = A_0 e^{-\lambda t} \qquad ---(15.27)\]
Compare Eq. (15.27) with Eq. (15.25). The activity falls off with exactly the same exponential factor as the number of nuclei, and with the same decay constant. This is extremely convenient in practice: measurements of activity, which a detector can make directly, can be used in place of counts of nuclei, which it cannot.
Units of Activity
Activity is measured in units of becquerel (Bq) in SI units. One becquerel is equal to one decay per second.
Another unit to measure activity is the curie (Ci). One curie is \(3.7 \times 10^{10}\) decays per second. Thus,
\[1 \text{ Ci} = 3.7 \times 10^{10} \text{ Bq}\]
The becquerel is a very small unit — a single decay per second — so practical activities are usually quoted with prefixes, in kBq, MBq, mCi or µCi.
Note the units carefully when solving such problems. If \(\lambda\) is given per day, then \(t\) must also be in days, and the activity obtained from \(A = \lambda N\) will be per day, not per second — a conversion that has to be made before quoting an answer in becquerel.
15.10.1 Half-life of Radioactive Material
Defining a Convenient Time Scale
The exponential law of Eq. (15.25) never gives a time at which a sample is fully decayed, so we need some other way of saying how quickly a species disappears. The usual measure is the time in which half of it is gone.
The time taken for the number of parent radioactive nuclei of a particular species to reduce to half its value is called the half-life \(T_{1/2}\) of the species.
This can be obtained from Eq. (15.25). Putting \(N = \dfrac{N_0}{2}\) at \(t = T_{1/2}\),
\[\frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}}\]
giving
\[e^{\lambda T_{1/2}} = 2\]
or
\[T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda} \qquad ---(15.28)\]
Here \(T_{1/2}\) is in second (s) if \(\lambda\) is in s\(^{-1}\); if \(\lambda\) is quoted per day or per year, the half-life comes out in the same unit of time.
Notice that \(N_0\) has cancelled out. The half-life does not depend on how much material we start with — a gram and a microgram of the same species both halve in exactly the same time. It is a fixed property of the species, just as \(\lambda\) is.
Diagram
Fig. 15.10b – Exponential decay curve with the vertical axis marked N₀, N₀/2, N₀/4 and N₀/8, and the corresponding times marked T₁/₂, 2T₁/₂ and 3T₁/₂ on the horizontal axis.
What Happens After the First Half-life
The interesting thing about half-life is that even though the number goes down from \(N_0\) to \(\dfrac{N_0}{2}\) in time \(T_{1/2}\), after another time interval \(T_{1/2}\) the number of parent nuclei will not go to zero. It will go to half of the value at \(t = T_{1/2}\), i.e., to \(\dfrac{N_0}{4}\).
Thus, in a time interval equal to the half-life, the number of parent nuclei reduces by a factor of \(\dfrac{1}{2}\), whatever the number happened to be at the start of that interval. The sequence therefore runs
\[N_0 \; \rightarrow \; \frac{N_0}{2} \; \rightarrow \; \frac{N_0}{4} \; \rightarrow \; \frac{N_0}{8} \; \rightarrow \; \ldots\]
after successive half-lives, as shown in Fig. 15.10b.
15.10.2 Average Life of a Radioactive Species
Averaging Over All the Nuclei
We have seen that different nuclei of a given radioactive species decay at different times, i.e., they have different life times. Some decay almost at once, others survive far longer, and there is no way of telling in advance which will do which. What we can do is work out the average of these life times over the whole sample.
We can calculate the average life time of a nucleus of the material using Eq. (15.25) as described below.
The number of nuclei decaying between time \(t\) and \(t + dt\) is given by \(\lambda N_0 e^{-\lambda t}\, dt\). The life time of these nuclei is \(t\), since they survived from the start until the instant \(t\).
To find the average, we multiply each life time by the number of nuclei having that life time, add over all the nuclei by integrating, and divide by the total number \(N_0\). Thus the average lifetime \(\tau\) of a nucleus is
\[\tau = \frac{1}{N_0}\int_{0}^{\infty} t \, \lambda N_0 e^{-\lambda t}\, dt\]
Integrating the above we get
\[\tau = \frac{1}{\lambda} \qquad ---(15.29)\]
with \(\tau\) in second (s) when \(\lambda\) is in s\(^{-1}\).
This result gives the decay constant a second, more tangible meaning. A species with \(\lambda = 0.01\) s\(^{-1}\) has nuclei that survive, on average, for 100 s.
The relation between the average life and half-life can be obtained using Eq. (15.28) as
\[T_{1/2} = \tau \ln 2 = 0.693 \, \tau \qquad ---(15.30)\]
Diagram
Fig. 15.10c – Exponential decay curve with both time scales marked: the half-life \(T_{1/2}\) at which N falls to N₀/2, and the average life \(\tau\) at which N falls to N₀/e, showing that \(\tau\) is the larger of the two.
15.11 Nuclear Energy
Why Another Source of Energy Is Needed
You are familiar with the naturally occurring, conventional sources of energy. These include the fossil fuels, i.e., coal, petroleum, natural gas, and firewood. The energy generation from these fuels is through chemical reactions — burning, in which atoms are rearranged into new molecules while their nuclei remain completely untouched.
It takes millions of years for these fuels to form. Naturally, the supply of these conventional sources is limited, and with indiscriminate use, they are bound to get over in a couple of hundred years from now.
Therefore, we have to use alternative sources of energy. The ones already in use are hydroelectric power, solar energy, wind energy and nuclear energy, nuclear energy being the largest source among these.
What Makes Nuclear Energy Different
Nuclear energy is the energy released when nuclei undergo a nuclear reaction, i.e., when one nucleus or a pair of nuclei, due to their interaction, undergo a change in their structure resulting in new nuclei and generating energy in the process.
The distinction from a chemical reaction is fundamental. In burning, electrons are shuffled between atoms and the nuclei are spectators. In a nuclear reaction, the nuclei themselves are rearranged, and the energies involved belong to an entirely different scale.
While the energy generated in chemical reactions is of the order of a few eV per reaction, the amount of energy released in a nuclear reaction is of the order of a few MeV. Thus, for the same weight of fuel, the nuclear energy released is about a million times that released through chemical reactions.
However, nuclear energy generation is a very complex and expensive process, and it can also be extremely harmful. The fuel is radioactive, the products of the reactions are radioactive, and both the reaction and the waste have to be kept under strict control. Let us learn more about it.
Where the Energy Comes From
We have seen in Section 15.8 that the mass of a nucleus is smaller than the sum of the masses of its constituents. The difference in these two masses is the binding energy of the nucleus.
It would be the energy released if the nucleus were formed by bringing together its constituents from infinity. This energy is large (in MeV), and this process can therefore be a good source of energy in principle.
In practice, we never form nuclei starting from individual nucleons. Free neutrons and protons in the required numbers are simply not available to be assembled at will, so the binding energy cannot be collected in this direct way.
However, we can obtain nuclear energy by two other processes:
nuclear fission, in which a heavy nucleus is broken into two nuclei of smaller masses, and
nuclear fusion, in which two light nuclei undergo a nuclear reaction and fuse together to form a heavier nucleus.
Both fission and fusion are nuclear reactions.
Reading Both Processes off One Graph
At first sight it seems contradictory that energy should be released both by splitting a nucleus and by joining nuclei together. The binding energy per nucleon curve of Fig. 15.6 resolves this at once, and it is worth fixing the idea before either process is studied in detail.
Energy is released whenever the products of a reaction are more tightly bound than the reactants — that is, whenever the reaction moves the material upwards on that curve, towards the peak near \(A = 56\). Heavy nuclei lie to the right of the peak, so splitting them into medium-mass fragments moves them up. Light nuclei lie to the left of the peak, so joining them into a heavier nucleus also moves them up, as indicated in Fig. 15.11a.
Diagram
Fig. 15.11a – The binding energy per nucleon curve with two arrows drawn on it: light nuclei on the left moving towards the peak by fusion, and heavy nuclei on the right moving towards the peak by fission.
In both cases the products together are lighter than what we began with, and the missing mass appears as energy, exactly as in the decays studied earlier. Let us understand how nuclear energy is released in the two processes.
15.11.1 Nuclear Fission
Why Splitting a Heavy Nucleus Releases Energy
We have seen in Fig. 15.6 that the binding energy per nucleon \(\left(\dfrac{E_B}{A}\right)\) depends on the mass number of the nuclei. This quantity is a measure of the stability of the nucleus.
As seen from the figure, the middle weight nuclei — with mass number ranging from 50 to 80 — have the highest binding energy per nucleon and are most stable, while nuclei with higher and lower atomic masses have smaller values of \(\dfrac{E_B}{A}\).
The value of \(\dfrac{E_B}{A}\) goes on decreasing till \(A \sim 238\), which is the mass number of the heaviest naturally occurring element, which is uranium. Many of the heavy nuclei are unstable and decay into two smaller mass nuclei.
Let us consider a case when a heavy nucleus, say with \(A \sim 230\), breaks into two nuclei having \(A\) between 50 and 150. The \(\dfrac{E_B}{A}\) of the product nuclei will be higher than that of the parent nucleus, since the products lie closer to the peak of the curve.
This means that the combined masses of the two product nuclei will be smaller than the mass of the parent nucleus — more tightly bound means lighter, by the mass-defect argument of Section 15.8. The difference in the mass of the parent nucleus and that of the product nuclei taken together will be released in the form of energy in the process.
This process, in which a heavy nucleus breaks into two lighter nuclei with the release of energy, is called nuclear fission, and is a source of nuclear energy.
Diagram
Fig. 15.11b – Nuclear fission: a slow neutron striking a heavy nucleus, the nucleus splitting into two fragments of unequal size, and two or three free neutrons flying outwards.
The Uranium Used in Fission
One of the nuclei used in nuclear energy generation by fission is \({}^{236}_{92}\text{U}\). This has a half-life of \(2.3 \times 10^{7}\) years and an activity of \(6.5 \times 10^{-5}\) Ci/g. However, it being fissionable, most of its nuclei have already decayed and it is not found in nature.
Natural uranium is a mixture of isotopes, and its composition matters a great deal here. More than 99% of natural uranium is in the form of \({}^{238}_{92}\text{U}\), and less than 1% is in the form of \({}^{235}_{92}\text{U}\).
\({}^{238}_{92}\text{U}\) also decays, but its half-life is about \(10^{3}\) times higher than that of \({}^{236}_{92}\text{U}\) and it is therefore not very useful for energy generation. A very long half-life means a very low rate of decay, so very little energy is released per second.
The species needed for nuclear energy generation, i.e. \({}^{236}_{92}\text{U}\), can be obtained from the naturally occurring \({}^{235}_{92}\text{U}\) by bombarding it with slow neutrons. \({}^{235}_{92}\text{U}\) absorbs a neutron and yields \({}^{236}_{92}\text{U}\). This reaction can be written as
\[{}^{235}_{92}\text{U} + n \rightarrow {}^{236}_{92}\text{U}\]
The Fission Reactions
\({}^{236}_{92}\text{U}\) can undergo fission in several ways, producing different pairs of daughter nuclei and generating different amounts of energy in the process. It does not split in one fixed manner; each nucleus takes one of many possible routes. Some of its decays are
\[{}^{236}_{92}\text{U} \rightarrow {}^{137}_{53}\text{I} + {}^{97}_{39}\text{Y} + 2n\] \[{}^{236}_{92}\text{U} \rightarrow {}^{140}_{56}\text{Ba} + {}^{94}_{36}\text{Kr} + 2n\] \[{}^{236}_{92}\text{U} \rightarrow {}^{133}_{51}\text{Sb} + {}^{99}_{41}\text{Nb} + 4n \qquad ---(15.31)\]
Check the bookkeeping in any one of these: the mass numbers and the atomic numbers each balance on the two sides, once the free neutrons are counted.
Some of the daughter nuclei produced are not stable and they further decay to produce more stable nuclei. These fission fragments are the source of the intense radioactivity of spent nuclear fuel, and of the difficulty of storing nuclear waste safely.
The energy produced in the fission is in the form of kinetic energy of the products, i.e., in the form of heat, which can be collected and converted to other forms of energy as needed. The fragments fly apart at high speed because they repel each other electrostatically, and they are quickly brought to rest by the surrounding material, which is thereby heated.
Chain Reaction
Neutrons are produced in the fission reactions shown in Eq. (15.31). This is the feature that makes fission useful as a source of power, because the reaction can then keep itself going.
Some reactions produce 2 neutrons while others produce 3 or 4. The average number of neutrons per reaction can be shown to be 2.7.
These neutrons are in turn absorbed by other \({}^{235}_{92}\text{U}\) nuclei to produce \({}^{236}_{92}\text{U}\), which undergo fission and produce a further 2.7 neutrons per fission.
This can have a cascading effect, and the number of neutrons produced — and therefore the number of \({}^{236}_{92}\text{U}\) nuclei produced — can increase quickly. This is called a chain reaction.
Diagram
Fig. 15.11c – Chain reaction: one neutron causing a fission that releases two neutrons, each of which causes a further fission, the number of events multiplying at each stage.
Such a reaction will lead to a fast increase in the number of fissions, and thereby a rapid increase in the amount of energy produced. Because each generation of neutrons takes only a tiny fraction of a second to produce the next, an uncontrolled multiplication builds up almost instantly. This will lead to an explosion.
Uranium Nuclear Reactor
In a nuclear reactor, methods are employed to stop a chain reaction from running away, and fission and energy generation are allowed to occur in a controlled fashion.
In a uranium reactor, \({}^{235}_{92}\text{U}\) is used as the fuel. It is bombarded by slow neutrons to produce \({}^{236}_{92}\text{U}\), which undergoes fission.
Two arrangements make the control possible. The neutrons released by fission are fast, so the fuel is surrounded by a material — a moderator — in which repeated collisions slow the neutrons down to the speeds at which they are readily absorbed by \({}^{235}_{92}\text{U}\). Alongside, rods of a strongly neutron-absorbing material are pushed into or drawn out of the core; pushing them in mops up surplus neutrons and slows the reaction, while withdrawing them allows it to build up. Between the two, the reaction is held at exactly one fission carrying on to the next.
The energy generated, which is in the form of heat, is carried away and converted to electricity by using turbines etc. In this last stage a nuclear station works like any thermal power station: the heat raises steam, the steam drives a turbine, and the turbine drives a generator. Only the source of the heat is different.
Diagram
Fig. 15.11d – Schematic of a nuclear reactor: fuel rods and control rods in a moderator-filled core, coolant carrying heat to a steam generator, and steam driving a turbine coupled to an electric generator.
More than 15 countries have nuclear reactors and use nuclear power. India is one of them. There are 22 nuclear reactors in India, the largest one being at Kudankulam, Tamil Nadu. Maximum nuclear power is generated by the USA.
15.11.2 Nuclear Fusion
Joining Light Nuclei
We have seen that light nuclei (\(A < 40\)) have lower \(\dfrac{E_B}{A}\) as compared to heavier ones. They sit on the rising part of the curve in Fig. 15.6, well to the left of the peak, so there is room for them to become more tightly bound.
If any two of the lighter nuclei come sufficiently close, within about one fm of each other, then they can undergo a nuclear reaction and form a heavier nucleus. One femtometre is the range within which the nuclear force acts, as described in Section 15.7.3.
The heavier nucleus will have higher \(\dfrac{E_B}{A}\) than the reactants. The mass of the product nucleus will therefore be lower than the total mass of the reactants, and energy of the order of MeV will be released in the process.
This process, wherein two nuclei fuse together to form a heavier nucleus accompanied by a release of nuclear energy, is called nuclear fusion.
Diagram
Fig. 15.11e – Nuclear fusion: two light nuclei approaching each other against their mutual repulsion, combining to form a single heavier nucleus, with energy released.
The Difficulty: Getting Two Nuclei Close Enough
For a nuclear reaction to take place, it is necessary for two nuclei to come to within about 1 fm of each other so that they can experience the nuclear forces. Everything difficult about fusion follows from this one requirement.
It is very difficult for two atoms to come that close to each other, due to the electrostatic repulsion between the electrons of the two atoms. Long before the nuclei can approach, the electron clouds meet and push the atoms apart.
This problem can be solved by stripping the atoms of their electrons and producing bare nuclei. It is possible to do so by giving the electrons energies larger than the ionization energies of the atoms, by heating a gas of atoms. The ionization energy was defined in Section 15.6.2; supplying more than this amount frees the electrons and leaves a gas of bare nuclei and free electrons.
But even after this, the two bare nuclei find it very difficult to go near each other, due to the repulsive force between their positive charges. This repulsion grows as the inverse square of the separation, so it becomes fiercest exactly where the nuclei need to arrive.
For nuclear fusion to occur, we have to heat the gas to a very high temperature, thereby providing the nuclei with very high kinetic energies. These high energies can help them to overcome the electrostatic repulsion and come close to one another.
As the positive charge of a nucleus goes on increasing with increase in its atomic number, the kinetic energies of the nuclei — i.e., the temperature of the gas necessary for nuclear fusion to occur — goes on increasing with increase in \(Z\). This is why fusion of hydrogen is the easiest case, and why heavier fuels demand hotter conditions still.
Fusion in the Sun
The temperature at the centre of the Sun is about \(10^{7}\) K. At such a temperature the material is fully ionised and the nuclei move fast enough for fusion to proceed.
The nuclear reactions taking place at the centre of the Sun are the fusion of four hydrogen nuclei, i.e., protons, to form a helium nucleus.
Of course, because of the electrostatic repulsion and the values of densities at the centre of the Sun, it is extremely unlikely that four protons will come sufficiently close to one another at a given time so that they can combine to form helium. A simultaneous four-body encounter is far too improbable.
Instead, the fusion proceeds in several steps, each involving only two particles at a time, with the intermediate products taking part in the next step. The effective reaction can be written as
\[4p \rightarrow \alpha + 2e^{+} + \text{neutrinos} + 26.7 \text{ MeV}\]
Diagram
Fig. 15.11f – The net solar fusion reaction: four protons combining in stages to give a helium nucleus, two positrons and neutrinos, with the energy released marked.
These reactions have been going on inside the Sun since past 4.5 billion years and are expected to continue for a similar time period in the future.
At the centres of other stars, where temperatures are higher, nuclei heavier than hydrogen can fuse, generating energy.
Fusion Throughout the Universe
Nuclear fusion is taking place all the time in the universe. It mostly takes place at the centres of stars, where the temperatures are high enough for nuclear reactions to take place. There, light nuclei fuse into heavier nuclei, generating energy. Nuclear fusion is in fact the source of energy for stars.
Most of the elements heavier than boron till iron, that we see around us today, have been produced through nuclear fusion inside stars. The material of the Earth, and of our own bodies, is largely the ash of fusion reactions that occurred in earlier generations of stars.
The Harmful Side of Nuclear Energy
The discussion on nuclear energy will not be complete without mentioning its harmful effects.
If an uncontrolled chain reaction sets up in a nuclear fuel, an extremely large amount of energy can be generated in a very short time. This is the same multiplication described in the previous section, with none of the control.
This fact has been used to produce what are called atom bombs or nuclear devices. Either fission alone, or both fission and fusion, are used in these bombs.
The first such devices were made towards the end of the second world war by America. By now, several countries including India have successfully made and tested such nuclear devices. America remains the only country to have actually used two atom bombs, which completely destroyed the cities of Hiroshima and Nagasaki in Japan in early August 1945.
15.12 Summary
Early Models of the Atom
The idea of indivisible atoms goes back to Leucippus and Democritus in the 5th century BC. Dalton’s theory stated that matter is made of indestructible particles, that atoms of a given element are identical, and that atoms combine to form new substances.
J. J. Thomson discovered the electron and showed that atoms are not indestructible. In his plum-pudding model (1903), the atom is a sphere of uniform positive charge with electrons embedded in it, the total positive charge equalling the total negative charge so that the atom is neutral. The model explained neutrality and the formation of ions, but not the way charge is actually distributed.
The Geiger-Marsden Experiment and Rutherford’s Model
Alpha particles (helium nuclei, charge \(+2e\)) from a collimated beam were fired at a thin gold foil, and the scattered particles were counted as scintillations through a movable microscope.
Most alpha particles passed almost undeflected; only about 0.14% were scattered through angles larger than 0.1°, about one in 8000 through angles larger than 90°, and a few through angles as large as 180°.
Rutherford concluded that the entire positive charge and about 99.9% of the mass of the atom are concentrated in a tiny central nucleus of size about 10 fm, which is about \(10^{-5}\) times the size of the atom and about \(10^{-15}\) times its volume, with electrons revolving around it in circular orbits.
Difficulties with Rutherford’s model: an electron in a circular orbit is accelerating, so by Maxwell’s theory it should radiate continuously, lose energy, spiral into the nucleus and cause the atom to collapse; and the frequency of the emitted radiation would change continuously. Neither is observed — atoms are stable and emit only certain frequencies, and only when supplied with energy.
Atomic Spectra
A heated solid gives a continuous spectrum, whereas a heated gas such as hydrogen gives a line spectrum consisting of a few sharp emission lines. Hydrogen emits at 410, 434, 486 and 656 nm in the visible region, with nothing in between.
The lines fall into series — Lyman, Balmer, Paschen, Brackett, Pfund — from shorter to longer wavelengths. Within each series the lines crowd closer together towards shorter wavelengths and approach a series limit.
The observed wavelengths obey
\[\frac{1}{\lambda} = R\left(\frac{1}{n^{2}} - \frac{1}{m^{2}}\right)\]
with \(n = 1, 2, 3, \ldots\) for the Lyman, Balmer, Paschen … series and \(m\) taking all integral values greater than \(n\). The series limit, obtained as \(m \rightarrow \infty\), is \(\lambda = \dfrac{n^{2}}{R}\).
- The wavelengths emitted are unique to each element, so an element can be identified from its spectrum.
Bohr’s Model
Postulate 1: electrons revolve around the nucleus in circular orbits, the centripetal force being provided by the electrostatic force of attraction between the electron and the nucleus.
Postulate 2: only those orbits are allowed in which the angular momentum is an integral multiple of \(\dfrac{h}{2\pi}\); electrons in these stable orbits do not radiate.
Postulate 3: an electron may make a transition to an orbit of lower energy, emitting a photon whose energy equals the difference of the two orbit energies.
The two basic equations are
\[m_e v_n r_n = \frac{nh}{2\pi} \qquad \text{and} \qquad \frac{m_e v_n^{2}}{r_n} = \frac{Ze^{2}}{4\pi \varepsilon_0 r_n^{2}}\]
- Solving them gives the radius and the speed in the \(n^{\text{th}}\) orbit:
\[r_n = \frac{\varepsilon_0 n^{2} h^{2}}{\pi m_e Z e^{2}} \qquad \text{and} \qquad v_n = \frac{Ze^{2}}{2 \varepsilon_0 n h}\]
- The radius is proportional to \(n^{2}\). For hydrogen (\(Z = 1\)) with \(n = 1\), \(r_1 = 0.053\) nm, called the Bohr radius \(a_0 = \dfrac{\varepsilon_0 h^{2}}{\pi m_e e^{2}}\), so that \(r_n = n^{2} a_0\).
Energy of the Electron
- The total energy of the electron in the \(n^{\text{th}}\) orbit is
\[E_n = -\frac{m_e Z^{2} e^{4}}{8 \varepsilon_0^{2} h^{2} n^{2}} = -13.6 \, \frac{Z^{2}}{n^{2}} \text{ eV}\]
The negative sign shows that the electron is bound; the energy becomes less negative as \(n\) increases and reaches zero as \(n \rightarrow \infty\).
\(n = 1\) is the ground state; higher values of \(n\) are excited states. For hydrogen the ground state energy is \(-13.6\) eV.
Excitation energy of a state is the energy needed to raise the electron to it from the ground state; for the \(n = 2\) state of hydrogen this is \(-3.4 - (-13.6) = 10.2\) eV.
Ionization energy is the minimum energy needed to set free an electron in the ground state; for hydrogen it is 13.6 eV, which is also its binding energy.
For a transition from the \(m^{\text{th}}\) to the \(n^{\text{th}}\) orbit,
\[\frac{1}{\lambda} = \frac{m_e Z^{2} e^{4}}{8 \varepsilon_0^{2} c h^{3}}\left(\frac{1}{n^{2}} - \frac{1}{m^{2}}\right) = R_H Z^{2}\left(\frac{1}{n^{2}} - \frac{1}{m^{2}}\right)\]
which is Rydberg’s formula, with \(R_H = \dfrac{m_e e^{4}}{8 \varepsilon_0^{2} c h^{3}} = 1.097 \times 10^{7}\) m\(^{-1}\). For hydrogen, \(Z = 1\).
Limitations of Bohr’s Model, and de Broglie’s Explanation
Bohr’s model could not explain the spectra of atoms other than hydrogen, nor the multiple components seen in some hydrogen lines; it gave no account of the differing intensities of the lines; and its quantisation condition was assumed without theoretical basis.
De Broglie treated the orbiting electron as a standing wave, requiring the circumference to contain a whole number of wavelengths:
\[2\pi r_n = n \lambda_n, \qquad \lambda_n = \frac{2\pi r_n}{n}\]
- With \(\lambda_n = \dfrac{h}{p_n} = \dfrac{h}{m v_n}\), this gives \(p_n = \dfrac{nh}{2\pi r_n}\) and hence \(L_n = p_n r_n = \dfrac{nh}{2\pi}\), which is Bohr’s second postulate.
The Nucleus
The nucleus contains protons and neutrons, together called nucleons. The proton mass is about 1836 times the electron mass, and the neutron is slightly heavier than the proton. The proton’s charge equals the electron’s in magnitude; the neutron is neutral.
\(Z\) is the atomic number (number of protons, and of electrons in the neutral atom), \(N\) the number of neutrons, and \(A = Z + N\) the mass number. A nuclide is written \({}^{A}_{Z}\text{X}\). Chemical properties are decided by \(Z\).
Isotopes have the same \(Z\) but different \(N\); isobars have the same \(A\); isotones have the same \(N\). Thus \({}^{2}_{1}\text{H}\) and \({}^{3}_{1}\text{H}\) are isotopes of hydrogen, \({}^{3}_{1}\text{H}\) and \({}^{3}_{2}\text{He}\) are isobars, and \({}^{3}_{1}\text{H}\) and \({}^{4}_{2}\text{He}\) are isotones.
Masses are measured in kg, in unified atomic mass units (1 u \(= 1.6605402 \times 10^{-27}\) kg, one-twelfth the mass of a neutral carbon-12 atom), or in energy-equivalent units, with 1 u \(= 931.5\) MeV/\(c^{2}\) from \(E = mc^{2}\). In these units \(m_e = 0.00055\) u, \(m_p = 1.007825\) u and \(m_n = 1.008665\) u.
Nuclear radius: \(R_X = R_0 A^{1/3}\) with \(R_0 = 1.2 \times 10^{-15}\) m, so nuclear volume is proportional to \(A\) and the density \(\rho = \dfrac{3m}{4\pi R_0^{3}}\) is the same for all nuclei, about \(2.3 \times 10^{17}\) kg m\(^{-3}\) — some thirteen orders of magnitude greater than the density of osmium, \(2.2 \times 10^{4}\) kg m\(^{-3}\).
The nuclear (strong) force binds nucleons together. It is the strongest force among subatomic particles, about 50–60 times stronger than the electrostatic force; it has a range of only a few fm, unlike the infinite range of electrostatic and gravitational forces; and it is charge independent.
Binding Energy
Mass defect: \(\Delta M = Z m_p + N m_n - M\), and binding energy \(E_B = \Delta M c^{2} = \left(Z m_p + N m_n - M\right)c^{2}\).
In terms of atomic masses, \(E_B = \left[Z m_H + N m_n - {}^{A}_{Z}M\right]c^{2}\), where \(m_H\) is the mass of a hydrogen atom.
Binding energy per nucleon \(\dfrac{E_B}{A}\) measures stability. It is least for deuterium, rises with \(A\), is highest on a plateau for \(A\) between 50 and 80 with a peak near \(A = 56\) (iron), and falls slowly beyond \(A = 80\). The binding energy of the hydrogen nucleus is zero.
Radioactive Decays
Unstable nuclei decay from a parent to a daughter nucleus. Radioactivity was discovered by Becquerel and is of three kinds.
Alpha decay: \({}^{A}_{Z}\text{X} \rightarrow {}^{A-4}_{Z-2}\text{Y} + \alpha\). All nuclei with \(A > 210\) undergo alpha decay, because electrostatic repulsion among the many protons outweighs the short-range nuclear attraction. Its energy release is \(Q = \left[m_X - m_Y - m_{He}\right]c^{2}\), appearing as kinetic energy of the products.
Beta minus decay: \(n \rightarrow p + e^{-} + \text{antineutrino}\), so \({}^{A}_{Z}\text{X} \rightarrow {}^{A}_{Z+1}\text{Y} + e^{-} + \text{antineutrino}\); \(A\) is unchanged and \(Z\) increases by one.
Beta plus decay: \(p \rightarrow n + e^{+} + \text{neutrino}\), so \({}^{A}_{Z}\text{X} \rightarrow {}^{A}_{Z-1}\text{Y} + e^{+} + \text{neutrino}\); \(A\) is unchanged and \(Z\) decreases by one. This cannot occur for a free proton, since the neutron is heavier, but can occur inside a nucleus. For beta decay, \(Q = \left[m_X - m_Y - m_e\right]c^{2}\).
Gamma decay: an excited nucleus emits a high energy photon; \(A\) and \(Z\) are unchanged. Nuclear energy levels are separated by keV to MeV, against a few eV for atomic levels, which is why nuclei emit gamma rays while atoms emit ultraviolet to radio waves. Gamma decay usually follows an alpha or beta decay.
A daughter nucleus may itself be unstable, giving a decay chain; \({}^{238}\text{U}\) decays 14 times by alpha and beta emission before reaching stable \({}^{206}\text{Pb}\).
Law of Radioactive Decay
The decay of an individual nucleus is unpredictable; for a large number, \(dN = -\lambda N(t)\, dt\), where \(\lambda\) is the decay constant.
Integration gives the decay law
\[N(t) = N_0 e^{-\lambda t}\]
- Activity is the number of decays per unit time:
\[A(t) = -\frac{dN}{dt} = \lambda N(t) = \lambda N_0 e^{-\lambda t} = A_0 e^{-\lambda t}\]
Units of activity: 1 becquerel (Bq) = 1 decay per second; 1 curie (Ci) \(= 3.7 \times 10^{10}\) Bq.
Half-life: \(T_{1/2} = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{\lambda}\). In each half-life the number of parent nuclei falls by a factor of \(\dfrac{1}{2}\), going \(N_0 \rightarrow \dfrac{N_0}{2} \rightarrow \dfrac{N_0}{4} \rightarrow \ldots\), never reaching zero.
Average life: \(\tau = \dfrac{1}{\lambda}\), related to half-life by \(T_{1/2} = \tau \ln 2 = 0.693\, \tau\).
Nuclear Energy
Chemical reactions release a few eV per reaction, nuclear reactions a few MeV, so for the same weight of fuel nuclear energy is about a million times greater. Nuclear energy generation is complex, expensive and potentially harmful.
Fission: a heavy nucleus splits into two lighter nuclei whose \(\dfrac{E_B}{A}\) is higher, so the products are lighter than the parent and energy is released. \({}^{235}_{92}\text{U}\) absorbs a slow neutron to give \({}^{236}_{92}\text{U}\), which fissions in several ways, for example
\[{}^{236}_{92}\text{U} \rightarrow {}^{137}_{53}\text{I} + {}^{97}_{39}\text{Y} + 2n\] \[{}^{236}_{92}\text{U} \rightarrow {}^{140}_{56}\text{Ba} + {}^{94}_{36}\text{Kr} + 2n\] \[{}^{236}_{92}\text{U} \rightarrow {}^{133}_{51}\text{Sb} + {}^{99}_{41}\text{Nb} + 4n\]
Natural uranium is more than 99% \({}^{238}\text{U}\) and less than 1% \({}^{235}\text{U}\). \({}^{236}\text{U}\) has a half-life of \(2.3 \times 10^{7}\) years and an activity of \(6.5 \times 10^{-5}\) Ci/g, and is not found in nature.
Each fission releases on average 2.7 neutrons, which can cause further fissions — a chain reaction. Uncontrolled, it leads to an explosion; in a nuclear reactor it is controlled so that heat is produced steadily and converted to electricity. India has 22 nuclear reactors, the largest at Kudankulam, Tamil Nadu, and the USA generates the most nuclear power.
Fusion: two light nuclei (\(A < 40\)) that come within about 1 fm of each other combine into a heavier nucleus of higher \(\dfrac{E_B}{A}\), releasing energy of the order of MeV. Very high temperatures are needed for the nuclei to overcome their electrostatic repulsion, and the required temperature rises with \(Z\).
At the centre of the Sun, at about \(10^{7}\) K, the effective reaction is
\[4p \rightarrow \alpha + 2e^{+} + \text{neutrinos} + 26.7 \text{ MeV}\]
proceeding in several steps and continuing for about 4.5 billion years so far. Fusion is the energy source of stars and has produced most elements heavier than boron up to iron.
15.13 Objective Questions
- Choose the correct option.
- In which of the following systems will the radius of the first orbit of the electron be smallest?
- hydrogen (B) singly ionized helium (C) deuteron (D) tritium
- The radius of the 4th orbit of the electron will be smaller than its 8th orbit by a factor of
- 2 (B) 4 (C) 8 (D) 16
- In the spectrum of hydrogen atom which transition will yield longest wavelength?
- \(n = 2\) to \(n = 1\) (B) \(n = 5\) to \(n = 4\) (C) \(n = 7\) to \(n = 6\) (D) \(n = 8\) to \(n = 7\)
- Which of the following properties of a nucleus does not depend on its mass number?
- radius (B) mass (C) volume (D) density
- If the number of nuclei in a radioactive sample at a given time is \(N\), what will be the number at the end of two half-lives?
- \(N/2\) (B) \(N/4\) (C) \(3N/4\) (D) \(N/8\)
15.14 Check Your Concepts
Thomson’s model explained why atoms emit electrons but never positively charged particles. Explain how the model accounted for this, and state why Rutherford’s model gives a different but equally satisfactory explanation of the same fact.
In the Geiger-Marsden experiment, only about one alpha particle in 8000 was deflected through more than 90°. Explain why this small fraction, rather than the large number that passed straight through, was the decisive evidence for a nucleus.
Why was a very thin foil essential in the alpha scattering experiment? What would be the difficulty in interpreting the results if a thick sheet of gold had been used?
An electron in Rutherford’s model moves with constant speed along a circular orbit. Explain why it is nevertheless said to be accelerating, and state the two separate consequences that classical electromagnetic theory predicts for such an electron.
A heated metal filament gives a continuous spectrum while heated hydrogen gas gives a line spectrum. Explain the reason for this difference in terms of the arrangement of atoms in the two cases.
Why do the successive lines within a given spectral series of hydrogen come closer and closer together as we move towards shorter wavelengths? Answer with reference to the Rydberg formula.
Bohr’s second postulate blocks the collapse of the atom predicted by classical physics. Explain how it does so, and state clearly what Bohr had to assume without proof.
The total energy of an electron in a Bohr orbit is negative. What is the physical meaning of this negative sign, and what would a total energy of zero represent?
Distinguish between the excitation energy of the \(n = 3\) state and the ionization energy of a hydrogen atom. Which of the two is larger, and why?
What happens to the energy levels of a one-electron system as \(Z\) increases? Using this, explain why the ionization energy of He\(^{+}\) is much greater than that of hydrogen.
Bohr’s model works well for hydrogen but fails for helium. Identify the assumption in the derivation that breaks down for a two-electron atom, and explain why.
Explain how treating the electron as a standing wave leads to Bohr’s condition on angular momentum. What would happen to the wave if the circumference of the orbit were not a whole number of wavelengths?
Two nuclides have the same mass number but different atomic numbers. What are such nuclides called, and which of their properties — chemical behaviour, nuclear radius, nuclear density — would you expect to be nearly the same for both?
The radius of a nucleus depends on its mass number but its density does not. Show, using \(R = R_0 A^{1/3}\), why the density is independent of \(A\), and state what this tells us about how closely nucleons are packed.
A student says that gravitational attraction between nucleons must be what holds the nucleus together, since gravity is always attractive. Explain why this reasoning is wrong.
State the three main properties of the nuclear force. For each one, state the corresponding property of the electrostatic force so that the contrast is clear.
Why is the binding energy per nucleon, rather than the total binding energy, used to compare the stabilities of different nuclei? Illustrate your answer by comparing a heavy nucleus with a medium-mass one.
The binding energy of the hydrogen nucleus is zero, while that of the deuterium nucleus is small but non-zero. Explain both facts.
Why do all nuclei with \(A > 210\) undergo alpha decay? Answer in terms of the range of the nuclear force and the range of the electrostatic force.
In beta-minus decay an electron is emitted from a nucleus, yet the nucleus contains no electrons. Explain where the emitted electron comes from and what happens to the mass number and atomic number of the nucleus.
Beta plus decay cannot occur for a free proton but can occur for a proton inside a nucleus. Explain this difference.
The beta particles emitted by a given radioactive species do not all have the same energy, although the \(Q\)-value of the decay is fixed. Explain this observation.
Why are the photons emitted by nuclei in the gamma region of the spectrum, while those emitted by atoms lie between the ultraviolet and the radio region?
Why does gamma decay usually occur immediately after an alpha or a beta decay, rather than on its own?
The decay of an individual nucleus is completely unpredictable, yet the decay of a sample follows a precise exponential law. Explain how both statements can be true at the same time.
Two samples contain equal numbers of nuclei of two different radioactive species. Sample P has a much larger decay constant than sample Q. Which sample has the greater initial activity, and which will still contain more undecayed nuclei after a long time? Give reasons.
A student states that after two half-lives a radioactive sample is completely decayed. Identify the error and state what fraction actually remains.
Why is the average life of a radioactive species always longer than its half-life? Give the relation between the two quantities.
Explain, using the binding energy per nucleon curve, why energy is released both when a heavy nucleus splits and when two light nuclei combine, although these are opposite processes.
Why must the neutrons used to induce fission in \({}^{235}\text{U}\) be slow, whereas fusion requires the reacting nuclei to be extremely fast?
Explain what a chain reaction is, and state what must be true of the number of neutrons from each fission that go on to cause further fissions if a reactor is to deliver steady power.
Natural uranium contains less than 1% of \({}^{235}\text{U}\), yet it is this isotope and not \({}^{238}\text{U}\) that is used as reactor fuel. Explain why.
Fusion releases more energy per kilogram of fuel than fission, yet fission reactors are in commercial use while fusion reactors are not. Explain the physical difficulty that makes controlled fusion so much harder to achieve.
Explain why fusion inside a star cannot continue past iron, and state where the elements heavier than iron are believed to have been formed.
Answer in brief.
- State the postulates of Bohr’s atomic model.
- State the difficulties faced by Rutherford’s atomic model.
- What are alpha, beta and gamma decays?
- Define excitation energy, binding energy and ionization energy of an electron in an atom.
- Show that the frequency of the first line in Lyman series is equal to the difference between the limiting frequencies of Lyman and Balmer series.
State the postulates of Bohr’s atomic model and derive the expression for the energy of an electron in the atom.
Starting from the formula for energy of an electron in the \(n^{\text{th}}\) orbit of hydrogen atom, derive the formula for the wavelengths of Lyman and Balmer series spectral lines and determine the shortest wavelengths of lines in both these series.
Determine the maximum angular speed of an electron moving in a stable orbit around the nucleus of hydrogen atom.
Determine the series limit of Balmer, Paschen and Bracket series, given the limit for Lyman series is 911.6 Å.
Describe alpha, beta and gamma decays and write down the formulae for the energies generated in each of these decays.
Explain what are nuclear fission and fusion giving an example of each. Write down the formulae for energy generated in each of these processes.
Describe the principles of a nuclear reactor. What is the difference between a nuclear reactor and a nuclear bomb?
15.15 Practice with Numericals
In solving problems, use \(m_e = 0.00055\) u \(= 0.5110\) MeV/\(c^{2}\), \(m_p = 1.00728\) u, \(m_n = 1.00866\) u, \(m_H = 1.007825\) u, u \(= 931.5\) MeV, \(e = 1.602 \times 10^{-19}\) C, \(h = 6.626 \times 10^{-34}\) J s, \(\varepsilon_0 = 8.854 \times 10^{-12}\) SI units and \(m_e = 9.109 \times 10^{-31}\) kg.
Calculate the radius of the 5th orbit of the electron in a hydrogen atom, given \(a_0 = 0.053\) nm. By what factor is it larger than the radius of the 2nd orbit?
The radius of a certain Bohr orbit of the hydrogen atom is 1.325 nm. Determine the principal quantum number of that orbit.
Calculate the radius of the first orbit of the electron in the Li\(^{2+}\) ion, for which \(Z = 3\). Compare it with the Bohr radius of hydrogen.
Determine the angular momentum of the electron in the 4th orbit of a hydrogen atom, and hence find the ratio of its angular momenta in the 4th and 2nd orbits.
Using \(v_n = \dfrac{Ze^{2}}{2\varepsilon_0 n h}\), calculate the speed of the electron in the first orbit of a hydrogen atom, and express it as a fraction of the speed of light.
Calculate the energy of the electron in the \(n = 4\) state of a hydrogen atom, and the energy needed to raise the electron to this state from the ground state.
An electron in a hydrogen atom makes a transition from the \(n = 4\) orbit to the \(n = 2\) orbit. Calculate (a) the energy of the emitted photon in eV, (b) its frequency, and (c) its wavelength. In which region of the spectrum does this line lie?
Calculate the wavelength of the first line of the Lyman series of hydrogen, taking \(R_H = 1.097 \times 10^{7}\) m\(^{-1}\).
Determine the shortest and the longest wavelengths of the Balmer series of hydrogen.
A photon of energy 12.75 eV is absorbed by a hydrogen atom in its ground state. To which energy level is the electron raised?
Calculate the ionization energy, in eV, of the He\(^{+}\) ion and of the Li\(^{2+}\) ion, and compare each with the ionization energy of hydrogen.
Calculate the de Broglie wavelength of the electron in the second orbit of a hydrogen atom, given \(r_2 = 0.212\) nm.
Calculate the radius of the \({}^{27}_{13}\text{Al}\) nucleus, taking \(R_0 = 1.2 \times 10^{-15}\) m. What is the ratio of the radii of the \({}^{27}_{13}\text{Al}\) and \({}^{125}_{52}\text{Te}\) nuclei?
A nucleus has a radius three times that of the \({}^{4}_{2}\text{He}\) nucleus. Determine its mass number.
Calculate the density of the \({}^{27}_{13}\text{Al}\) nucleus, taking the mass of a nucleon as \(1.67 \times 10^{-27}\) kg, and verify that the value obtained is of the order given in the chapter.
Calculate the electrostatic force of repulsion between two protons separated by \(2.0 \times 10^{-15}\) m, and state what this tells you about the strength required of the nuclear force at that separation.
The atomic mass of \({}^{12}_{6}\text{C}\) is 12.000000 u. Calculate its mass defect, its binding energy, and its binding energy per nucleon.
The atomic mass of \({}^{56}_{26}\text{Fe}\) is 55.934939 u. Determine its binding energy per nucleon and compare your value with the position of the peak in the binding energy curve.
The atomic mass of \({}^{2}_{1}\text{H}\) is 2.014102 u. Calculate the binding energy of the deuterium nucleus and its binding energy per nucleon.
A nucleus \({}^{210}_{84}\text{Po}\) undergoes alpha decay. Write the decay equation, identifying the daughter nucleus, and calculate the \(Q\)-value given the atomic masses \({}^{210}_{84}\text{Po} = 209.982874\) u, \({}^{206}_{82}\text{Pb} = 205.974465\) u and \({}^{4}_{2}\text{He} = 4.002603\) u.
Write the beta-minus decay equation for \({}^{14}_{6}\text{C}\) and identify the daughter nucleus. Calculate the \(Q\)-value of the decay given the atomic masses \({}^{14}_{6}\text{C} = 14.003242\) u and \({}^{14}_{7}\text{N} = 14.003074\) u.
An excited nucleus emits a gamma ray of energy 0.662 MeV. Calculate the frequency and the wavelength of this photon.
A radioactive species has a half-life of 12.0 hours. Calculate its decay constant and its average life.
A sample initially contains \(8.0 \times 10^{18}\) nuclei of a species whose half-life is 5.0 days. Calculate the number of undecayed nuclei and the activity of the sample after 15.0 days.
The activity of a sample falls from \(6.4 \times 10^{5}\) Bq to \(4.0 \times 10^{4}\) Bq in 32 minutes. Determine its half-life and its decay constant.
A radioactive sample has an activity of 3.7 kBq. Express this activity in curie, and calculate the number of nuclei present if the decay constant of the species is \(1.5 \times 10^{-9}\) s\(^{-1}\).
What fraction of a radioactive sample remains undecayed after a time equal to (a) its average life and (b) three times its half-life?
A radioactive isotope used in medicine has a half-life of 6.0 hours. If a hospital receives a source of activity 500 MBq, what will its activity be 24 hours later?
Calculate the energy released in the fission reaction \({}^{236}_{92}\text{U} \rightarrow {}^{133}_{51}\text{Sb} + {}^{99}_{41}\text{Nb} + 4n\), given the atomic masses \({}^{236}_{92}\text{U} = 236.04557\) u, \({}^{133}_{51}\text{Sb} = 132.91525\) u and \({}^{99}_{41}\text{Nb} = 98.91163\) u.
A reactor generates 500 MW of thermal power. Taking the average energy released per fission as 200 MeV, calculate the number of fissions occurring per second and the mass of \({}^{235}\text{U}\) consumed in one day.
Calculate the energy released in the fusion reaction \({}^{2}_{1}\text{H} + {}^{3}_{1}\text{H} \rightarrow {}^{4}_{2}\text{He} + n\), given the atomic masses \({}^{2}_{1}\text{H} = 2.014102\) u, \({}^{3}_{1}\text{H} = 3.016049\) u and \({}^{4}_{2}\text{He} = 4.002603\) u.
The Sun radiates energy at the rate of about \(3.8 \times 10^{26}\) W. Taking the effective fusion reaction to release 26.7 MeV, calculate the number of such reactions occurring per second inside the Sun.
Calculate the binding energy of an alpha particle given its mass to be 4.00151 u.
An electron in hydrogen atom stays in its second orbit for \(10^{-8}\) s. How many revolutions will it make around the nucleus in that time?
Determine the binding energy per nucleon of the americium isotope \({}^{244}_{95}\text{Am}\), given the mass of \({}^{244}_{95}\text{Am}\) to be 244.06428 u.
Calculate the energy released in the nuclear reaction \({}^{7}_{3}\text{Li} + p \rightarrow 2\alpha\) given mass of \({}^{7}_{3}\text{Li}\) atom and of helium atom to be 7.016 u and 4.0026 u respectively.
Complete the following equations describing nuclear decays.
- \({}^{226}_{88}\text{Ra} \rightarrow \alpha + \underline{\hspace{2cm}}\)
- \({}^{19}_{8}\text{O} \rightarrow \underline{\hspace{2cm}} + e^{-}\)
- \({}^{228}_{90}\text{Th} \rightarrow \alpha + \underline{\hspace{2cm}}\)
- \({}^{12}_{7}\text{N} \rightarrow {}^{12}_{6}\text{C} + \underline{\hspace{2cm}}\)
- Calculate the energy released in the following reactions, given the masses to be \({}^{223}_{88}\text{Ra}\): 223.0185 u, \({}^{209}_{82}\text{Pb}\): 208.9811, \({}^{14}_{6}\text{C}\): 14.00324, \({}^{236}_{92}\text{U}\): 236.0456, \({}^{140}_{56}\text{Ba}\): 139.9106, \({}^{94}_{36}\text{Kr}\): 93.9341, \({}^{11}_{6}\text{C}\): 11.01143, \({}^{11}_{5}\text{B}\): 11.0093. Ignore neutrino energy.
- \({}^{223}_{88}\text{Ra} \rightarrow {}^{209}_{82}\text{Pb} + {}^{14}_{6}\text{C}\)
- \({}^{236}_{92}\text{U} \rightarrow {}^{140}_{56}\text{Ba} + {}^{94}_{36}\text{Kr} + 2n\)
- \({}^{11}_{6}\text{C} \rightarrow {}^{11}_{5}\text{B} + e^{+} + \text{neutrino}\)
Sample of carbon obtained from any living organism has a decay rate of 15.3 decays per gram per minute. A sample of carbon obtained from very old charcoal shows a disintegration rate of 12.3 disintegrations per gram per minute. Determine the age of the old sample given the decay constant of carbon to be \(3.839 \times 10^{-12}\) per second.
The half-life of \({}^{90}_{38}\text{Sr}\) is 28 years. Determine the disintegration rate of its 5 mg sample.
What is the amount of \({}^{60}_{27}\text{Co}\) necessary to provide a radioactive source of strength 10.0 mCi, its half-life being 5.3 years?
Disintegration rate of a sample is \(10^{10}\) per hour at 20 hrs from the start. It reduces to \(6.3 \times 10^{9}\) per hour after 30 hours. Calculate its half life and the initial number of radioactive atoms in the sample.
The isotope \({}^{57}\text{Co}\) decays by electron capture to \({}^{57}\text{Fe}\) with a half-life of 272 d. The \({}^{57}\text{Fe}\) nucleus is produced in an excited state, and it almost instantaneously emits gamma rays. (a) Find the mean lifetime and decay constant for \({}^{57}\text{Co}\). (b) If the activity of a radiation source \({}^{57}\text{Co}\) is 2.0 µCi now, how many \({}^{57}\text{Co}\) nuclei does the source contain? (c) What will be the activity after one year?
A source contains two species of phosphorous nuclei, \({}^{32}_{15}\text{P}\) (\(T_{1/2} = 14.3\) d) and \({}^{33}_{15}\text{P}\) (\(T_{1/2} = 25.3\) d). At time \(t = 0\), 90% of the decays are from \({}^{32}_{15}\text{P}\). How much time has to elapse for only 15% of the decays to be from \({}^{32}_{15}\text{P}\)?
Before the year 1900 the activity per unit mass of atmospheric carbon due to the presence of \({}^{14}\text{C}\) averaged about 0.255 Bq per gram of carbon. (a) What fraction of carbon atoms were \({}^{14}\text{C}\)? (b) An archaeological specimen containing 500 mg of carbon, shows 174 decays in one hour. What is the age of the specimen, assuming that its activity per unit mass of carbon when the specimen died was equal to the average value of the air? Half-life of \({}^{14}\text{C}\) is 5730 years?
How much mass of \({}^{235}\text{U}\) is required to undergo fission each day to provide 3000 MW of thermal power? Average energy per fission is 202.79 MeV.
In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are \({}^{24}_{12}\text{Mg}\) (23.98504 u), \({}^{25}_{12}\text{Mg}\) (24.98584 u) and \({}^{26}_{12}\text{Mg}\) (25.98259 u). The natural abundance of \({}^{24}_{12}\text{Mg}\) is 78.99% by mass. Calculate the abundances of other two isotopes.