13.7 Nuclear Energy

The curve of binding energy per nucleon \(E_{b n}\), given in Fig. 13.1, has a long flat middle region between \(A=30\) and \(A=170\). In this region the binding energy per nucleon is nearly constant ( 8.0 MeV ). For the lighter nuclei region, \(A<30\), and for the heavier nuclei region, \(A>170\), the binding energy per nucleon is less than 8.0 MeV, as we have noted earlier. Now, the greater the binding energy, the less is the total mass of a bound system, such as a nucleus. Consequently, if nuclei with less total binding energy transform to nuclei with greater binding energy, there will be a net energy release. This is what happens when a heavy nucleus decays into two or more intermediate mass fragments (fission) or when light nuclei fuse into a havier nucleus (fusion.)

Exothermic chemical reactions underlie conventional energy sources such as coal or petroleum. Here the energies involved are in the range of electron volts. On the other hand, in a nuclear reaction, the energy release is of the order of MeV. Thus for the same quantity of matter, nuclear sources produce a million times more energy than a chemical source. Fission of 1 kg of uranium, for example, generates \(10^{14} \mathrm{~J}\) of energy; compare it with burning of 1 kg of coal that gives \(10^7 \mathrm{~J}\).

13.7.1 Fission

New possibilities emerge when we go beyond natural radioactive decays and study nuclear reactions by bombarding nuclei with other nuclear particles such as proton, neutron, \(\alpha\)-particle, etc.

A most important neutron-induced nuclear reaction is fission. An example of fission is when a uranium isotope \({ }_{92}^{235} \mathrm{U}\) bombarded with a neutron breaks into two intermediate mass nuclear fragments
\(
{ }_0^1 \mathrm{n}+{ }_{92}^{235} \mathrm{U} \rightarrow{ }_{92}^{236} \mathrm{U} \rightarrow{ }_{56}^{144} \mathrm{Ba}+{ }_{36}^{89} \mathrm{Kr}+3{ }_0^1 \mathrm{n} \dots(13.10)
\)
The same reaction can produce other pairs of intermediate mass fragments
\(
{ }_0^1 \mathrm{n}+{ }_{92}^{235} \mathrm{U} \rightarrow{ }_{92}^{236} \mathrm{U} \rightarrow{ }_{51}^{133} \mathrm{Sb}+{ }_{41}^{99} \mathrm{Nb}+4{ }_0^1 \mathrm{n} \dots(13.11)
\)
\(
\begin{aligned}
&\text { Or, as another example, }\\
&{ }_0^1 \mathrm{n}+{ }_{92}^{235} \mathrm{U} \rightarrow{ }_{54}^{140} \mathrm{Xe}+{ }_{38}^{94} \mathrm{Sr}+2{ }_0^1 \mathrm{n} \dots(13.12)
\end{aligned}
\)
The fragment products are radioactive nuclei; they emit \(\beta\) particles in succession to achieve stable end products.

The energy released (the \(Q\) value ) in the fission reaction of nuclei like uranium is of the order of 200 MeV per fissioning nucleus. This is estimated as follows:

Let us take a nucleus with \(A=240\) breaking into two fragments each of \(A=120\). Then
\(E_{b n}\) for \(A=240\) nucleus is about 7.6 MeV,
\(E_{b n}\) for the two \(A=120\) fragment nuclei is about 8.5 MeV.
\(\therefore\) Gain in binding energy for nucleon is about 0.9 MeV.
Hence the total gain in binding energy is \(240 \times 0.9\) or 216 MeV.
The disintegration energy in fission events first appears as the kinetic energy of the fragments and neutrons. Eventually it is transferred to the surrounding matter appearing as heat. The source of energy in nuclear reactors, which produce electricity, is nuclear fission. The enormous energy released in an atom bomb comes from uncontrolled nuclear fission.

13.7.2 Nuclear fusion – energy generation in stars

When two light nuclei fuse to form a larger nucleus, energy is released, since the larger nucleus is more tightly bound, as seen from the binding energy curve in Fig. 13.1. Some examples of such energy-liberating nuclear fusion reactions are :
\(
\begin{aligned}
& { }_1^1 \mathrm{H}+{ }_1^1 \mathrm{H} \rightarrow{ }_1^2 \mathrm{H}+e^{+}+\nu+0.42 \mathrm{MeV} \dots[13.13(a)]\\
& { }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H} \rightarrow{ }_2^3 \mathrm{He}+n+3.27 \mathrm{MeV} \dots[13.13(b)]\\
& { }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H} \rightarrow{ }_1^3 \mathrm{H}+{ }_1^1 \mathrm{H}+4.03 \mathrm{MeV} \dots[13.13(c)]
\end{aligned}
\)
In the first reaction, two protons combine to form a deuteron and a positron with a release of 0.42 MeV energy. In reaction [13.13(b)], two deuterons combine to form the light isotope of helium. In reaction (13.13c), two deuterons combine to form a triton and a proton. For fusion to take place, the two nuclei must come close enough so that attractive short-range nuclear force is able to affect them. However, since they are both positively charged particles, they experience coulomb repulsion. They, therefore, must have enough energy to overcome this coulomb barrier. The height of the barrier depends on the charges and radii of the two interacting nuclei. It can be shown, for example, that the barrier height for two protons is \(\sim 400 \mathrm{keV}\), and is higher for nuclei with higher charges. We can estimate the temperature at which two protons in a proton gas would (averagely) have enough energy to overcome the coulomb barrier:

\((3 / 2) k T=K \simeq 400 \mathrm{keV}\), which gives \(\mathrm{T} \sim 3 \times 10^9 \mathrm{~K}\).
When fusion is achieved by raising the temperature of the system so that particles have enough kinetic energy to overcome the coulomb repulsive behaviour, it is called thermonuclear fusion.

Thermonuclear fusion is the source of energy output in the interior of stars. The interior of the sun has a temperature of \(1.5 \times 10^7 \mathrm{~K}\), which is considerably less than the estimated temperature required for fusion of particles of average energy. Clearly, fusion in the sun involves protons whose energies are much above the average energy.

The fusion reaction in the sun is a multi-step process in which the hydrogen is burned into helium. Thus, the fuel in the sun is the hydrogen in its core. The proton-proton ( \(\mathrm{p}, \mathrm{p}\) ) cycle by which this occurs is represented by the following sets of reactions:
\(
\begin{aligned}
& { }_1^1 \mathrm{H}+{ }_1^1 \mathrm{H} \rightarrow{ }_1^2 \mathrm{H}+e^{+}+\nu+0.42 \mathrm{MeV} \dots(i)\\
& e^{+}+e^{-} \rightarrow \gamma+\gamma+1.02 \mathrm{MeV} \dots(ii)\\
& { }_1^2 \mathrm{H}+{ }_1^1 \mathrm{H} \rightarrow{ }_2^3 \mathrm{He}+\gamma+5.49 \mathrm{MeV} \dots(iii)\\
& { }_2^3 \mathrm{He}+{ }_2^3 \mathrm{He} \rightarrow{ }_2^4 \mathrm{He}+{ }_1^1 \mathrm{H}+{ }_1^1 \mathrm{H}+12.86 \mathrm{MeV} \dots(iv) (13.14)
\end{aligned}
\)
For the fourth reaction to occur, the first three reactions must occur twice, in which case two light helium nuclei unite to form ordinary helium nucleus. If we consider the combination 2 (i) +2 (ii) +2 (iii) + (iv), the net effect is
\(
4{ }_1^1 \mathrm{H}+2 e^{-} \rightarrow{ }_2^4 \mathrm{He}+2 \nu+6 \gamma+26.7 \mathrm{MeV}
\)
\(
\text { or }\left(4_1^1 \mathrm{H}+4 e^{-}\right) \rightarrow\left({ }_2^4 \mathrm{He}+2 e^{-}\right)+2 \nu+6 \gamma+26.7 \mathrm{MeV} \dots(13.15)
\)
Thus, four hydrogen atoms combine to form an \({ }_2^4 \mathrm{He}\) atom with a release of 26.7 MeV of energy.

Helium is not the only element that can be synthesized in the interior of a star. As the hydrogen in the core gets depleted and becomes helium, the core starts to cool. The star begins to collapse under its own gravity which increases the temperature of the core. If this temperature increases to about \(10^8 \mathrm{~K}\), fusion takes place again, this time of helium nuclei into carbon. This kind of process can generate through fusion higher and higher mass number elements. But elements more massive than those near the peak of the binding energy curve in Fig. 13.1 cannot be so produced.

The age of the sun is about \(5 \times 10^9 \mathrm{y}\) and it is estimated that there is enough hydrogen in the sun to keep it going for another 5 billion years. After that, the hydrogen burning will stop and the sun will begin to cool and will start to collapse under gravity, which will raise the core temperature. The outer envelope of the sun will expand, turning it into the so called red giant.

13.7.3 Controlled thermonuclear fusion

The natural thermonuclear fusion process in a star is replicated in a thermonuclear fusion device. In controlled fusion reactors, the aim is to generate steady power by heating the nuclear fuel to a temperature in the range of \(10^8 \mathrm{~K}\). At these temperatures, the fuel is a mixture of positive ions and electrons (plasma). The challenge is to confine this plasma since no container can stand such a high temperature. Several countries around the world including India are developing techniques in this connection. If successful, fusion reactors will hopefully supply almost unlimited power to humanity.

Example 13.4: Answer the following questions:
(a) Are the equations of nuclear reactions (such as those given in Section 13.7) ‘balanced’ in the sense a chemical equation (e.g., \(2 \mathrm{H}_2+\mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}\) ) is? If not, in what sense are they balanced on both sides?
(b) If both the number of protons and the number of neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice-versa) in a nuclear reaction?
(c) A general impression exists that mass-energy interconversion takes place only in nuclear reaction and never in chemical reaction. This is strictly speaking, incorrect. Explain.

Solution: (a) A chemical equation is balanced in the sense that the number of atoms of each element is the same on both sides of the equation. A chemical reaction merely alters the original combinations of atoms. In a nuclear reaction, elements may be transmuted. Thus, the number of atoms of each element is not necessarily conserved in a nuclear reaction. However, the number of protons and the number of neutrons are both separately conserved in a nuclear reaction. [Actually, even this is not strictly true in the realm of very high energies – what is strictly conserved is the total charge and total ‘baryon number’. We need not pursue this matter here.] In nuclear reactions (e.g., Eq. 13.10), the number of protons and the number of neutrons are the same on the two sides of the equation.
(b) We know that the binding energy of a nucleus gives a negative contribution to the mass of the nucleus (mass defect). Now, since proton number and neutron number are conserved in a nuclear reaction, the total rest mass of neutrons and protons is the same on either side of a reaction. But the total binding energy of nuclei on the left side need not be the same as that on the right hand side. The difference in these binding energies appears as energy released or absorbed in a nuclear reaction. Since binding energy contributes to mass, we say that the difference in the total mass of nuclei on the two sides get converted into energy or vice-versa. It is in these sense that a nuclear reaction is an example of mass-energy interconversion.
(c) From the point of view of mass-energy interconversion, a chemical reaction is similar to a nuclear reaction in principle. The energy released or absorbed in a chemical reaction can be traced to the difference in chemical (not nuclear) binding energies of atoms and molecules on the two sides of a reaction. Since, strictly speaking, chemical binding energy also gives a negative contribution (mass defect) to the total mass of an atom or molecule, we can equally well say that the difference in the total mass of atoms or molecules, on the two sides of the chemical reaction gets converted into energy or vice-versa. However, the mass defects involved in a chemical reaction are almost a million times smaller than those in a nuclear reaction. This is the reason for the general impression, (which is incorrect) that mass-energy interconversion does not take place in a chemical reaction.