Theory, worked examples and 15 previous-year questions with full solutions — IIT JAM, GATE, JEST, TIFR and CSIR NET-JRF, 2020 onwards.
Theory
Chapter 4 treated decay through the decay constant \(\lambda\) without asking what fixes its value. Here we study the three classical modes one by one. In \(\alpha\) decay a heavy nucleus emits a \({}^4\mathrm{He}\) nucleus and escapes from the nucleus by tunnelling through the Coulomb barrier; in \(\beta\) decay a neutron turns into a proton, or a proton into a neutron, through the weak interaction, with the emission of a neutrino; in \(\gamma\) decay an excited nucleus emits a photon, or ejects an atomic electron by internal conversion, subject to multipole selection rules that make \(\gamma\) spectroscopy the main tool for finding nuclear spins and parities.
5.1 Alpha decay: energetics
(a) The Q-value
In \(\alpha\) decay \({}^{A}_{Z}\mathrm{X}\to{}^{A-4}_{Z-2}\mathrm{Y} + {}^{4}_{2}\mathrm{He}\), for example \({}^{238}\mathrm{U}\to{}^{234}\mathrm{Th}+\alpha\). Alpha emitters are almost all heavier than lead; their \(\alpha\) energies lie in the narrow range 4–9 MeV while their half-lives range from \(10^{-7}\) s to more than \(10^{17}\) y. For the parent at rest, energy conservation gives the disintegration energy\[ \boxed{\,Q_\alpha = (M_X - M_Y - M_\alpha)c^2 = K_\alpha + K_Y = B(A-4,Z-2) + B({}^4\mathrm{He}) - B(A,Z)\,} \] with \(B({}^4\mathrm{He}) = 28.3\) MeV. Atomic masses may be used, since the electron numbers balance (\(M({}^4\mathrm{He}) = 4.002603\) u). The decay is possible only if \(Q_\alpha>0\). Removing four nucleons from a heavy nucleus costs about \(4\times7.6\approx30\) MeV, but forming the \(\alpha\) particle returns 28.3 MeV, and because \(B/A\) falls with \(A\) for heavy nuclei, \(Q_\alpha\) becomes positive for \(A\gtrsim150\) and reaches about 5 MeV near uranium. The exceptional binding of \({}^4\mathrm{He}\) is also why heavy nuclei emit \(\alpha\) particles rather than single nucleons, deuterons or \({}^3\mathrm{He}\), for which \(Q<0\).
(b) Kinetic energy of the alpha particle and recoil
The \(\alpha\) particle and the daughter fly apart with equal and opposite momenta, so their non-relativistic kinetic energies \(p^2/2M\) are shared in inverse proportion to the masses. With \(K_\alpha + K_Y = Q_\alpha\), \[ \boxed{\,K_\alpha = \frac{M_Y}{M_Y + M_\alpha}\,Q_\alpha\approx\frac{A-4}{A}\,Q_\alpha,\qquad K_Y = \frac{M_\alpha}{M_Y+M_\alpha}\,Q_\alpha\approx\frac{4}{A}\,Q_\alpha\,} \] where \(A\) is the mass number of the parent. For \(A\approx230\) the daughter recoils with about 2 per cent of \(Q_\alpha\), of order 0.1 MeV. Since the decay is two-body, the \(\alpha\) particles of a given transition are monoenergetic. Many emitters show fine structure: several groups of \(\alpha\) particles, the most energetic leading to the ground state of the daughter and the others to excited states that then emit \(\gamma\) rays. A state at excitation \(E^*\) has \(Q = Q_0 - E^*\), so \(E^* = (K_{\alpha,0} - K_{\alpha})A/(A-4)\); the maximum \(\alpha\) energy belongs to the ground-state transition. Monoenergetic \(\alpha\) particles have a well-defined range, a few centimetres in air; empirically \(R\propto v^3\propto K^{3/2}\), with \(R_{\rm air}\approx0.318\,K^{3/2}\) cm for \(K\) in MeV.
Example 5.1 The atomic masses of \({}^{226}\mathrm{Ra}\), \({}^{222}\mathrm{Rn}\) and \({}^{4}\mathrm{He}\) are 226.025410 u, 222.017578 u and 4.002603 u. Find \(Q_\alpha\), \(K_\alpha\) and the recoil energy of the radon. A second, weaker \(\alpha\) group has 4.601 MeV; find the energy of the accompanying \(\gamma\) ray.
Solution.\(Q_\alpha = 0.005229\times931.494\) MeV \(= 4.871\) MeV, so \(K_\alpha = 4.871\times222/226 = 4.785\) MeV and \(K_{\rm Rn} = 4\times4.871/226 = 0.086\) MeV. The two groups differ in \(Q\) by \((4.785 - 4.601)\times226/222 = 0.186\) MeV, so the weaker group populates a level at 186 keV, which emits a 186 keV \(\gamma\) ray. The bare difference of \(\alpha\) energies would underestimate it, because the daughter recoil is ignored.
5.2 Theory of alpha decay
(a) The Coulomb barrier
Beyond the range of the nuclear force the \(\alpha\) particle and the daughter (charge \(Z_de\), \(Z_d = Z-2\)) repel with \(V_C(r) = 2Z_de^2/(4\pi\varepsilon_0r)\). The repulsion is greatest at the point of separation, where the two nuclei just touch, giving the height of the Coulomb barrier\[ \boxed{\,R = r_0\left(A_d^{1/3} + A_\alpha^{1/3}\right),\qquad V_C(R) = \frac{2Z_de^2}{4\pi\varepsilon_0R}\,} \] Inside \(R\) the potential is a deep attractive well (Figure 5.1). For heavy nuclei \(R\approx9\)–10 fm and \(V_C(R)\approx25\)–28 MeV, several times the 4–9 MeV with which \(\alpha\) particles emerge, so the \(\alpha\) particle cannot escape classically. An \(\alpha\) particle of energy \(Q_\alpha\) meets the barrier at the classical turning point\(b\), and the forbidden region \(R<r<b\) is the barrier, of width \(b - R\): \[ \boxed{\,b = \frac{2Z_de^2}{4\pi\varepsilon_0Q_\alpha}\,} \]
Figure 5.1 Potential energy of the \(\alpha\) particle and daughter against separation \(r\). The \(\alpha\) particle of energy \(Q\) tunnels through the barrier between \(R\) and the turning point \(b\).
(b) Gamow’s theory
Gamow, and Gurney and Condon (1928), assumed that an \(\alpha\) particle preformed in the nucleus strikes the barrier at a frequency \(\nu\approx v/2R\) (about \(10^{21}\) s\(^{-1}\)) and tunnels through it with probability \(T = e^{-2G}\), so that \[ \boxed{\,\lambda = P\,\nu\,e^{-2G},\qquad G = \frac{1}{\hbar}\int_R^b\sqrt{2m\,[V_C(r) - Q_\alpha]}\;dr\,} \] where \(P\le1\) is the preformation probability and \(m\) the reduced mass. With \(V_C = Q_\alpha b/r\) and the substitution \(r = b\cos^2\theta\) the integral is elementary: \[ G = \frac{\sqrt{2mQ_\alpha}}{\hbar}\,b\left[\arccos\sqrt{x} - \sqrt{x(1-x)}\right],\qquad x = \frac{R}{b}. \] For a thick barrier (\(R\ll b\)) the bracket is \(\pi/2 - 2\sqrt{R/b}\), and with \(e^2/4\pi\varepsilon_0 = 1.44\) MeV fm and \(m_\alpha c^2 = 3727\) MeV, \[ \boxed{\begin{aligned} 2G&\approx\frac{4\pi Z_de^2}{4\pi\varepsilon_0\hbar v} - \frac{8}{\hbar}\sqrt{\frac{Z_de^2mR}{4\pi\varepsilon_0}}\\ &\approx3.96\,\frac{Z_d}{\sqrt{Q_\alpha\,[\text{MeV}]}} - 2.97\sqrt{Z_d\,R\,[\text{fm}]} \end{aligned}} \] For \({}^{238}\mathrm{U}\) (\(Z_d = 90\), \(Q_\alpha = 4.27\) MeV, \(R = 9.3\) fm) \(2G\approx88\) and \(T\approx10^{-38}\); with \(\nu\approx8\times10^{20}\) s\(^{-1}\) this gives \(t_{1/2}\approx5\times10^{9}\) y, close to the measured value. Because \(2G\propto Z_d/\sqrt{Q_\alpha}\) appears in an exponent of order 30–90, doubling \(Q_\alpha\) changes the half-life by some 20 orders of magnitude.
(c) The Geiger–Nuttall law
Taking logarithms of \(t_{1/2} = \ln2/(P\nu e^{-2G})\), with \(\nu\) and the second term of \(2G\) slowly varying, gives the empirical Geiger–Nuttall law\[ \boxed{\,\log_{10}t_{1/2} = a + b\,\frac{Z}{\sqrt{Q_\alpha}}\,} \] a straight line for the isotopes of each element: shorter-lived emitters emit more energetic \(\alpha\) particles, and at equal \(Q_\alpha\) a higher \(Z\) (higher barrier) gives a longer half-life.
(d) Angular momentum and parity
The \(\alpha\) particle is \(0^+\), so it carries only orbital angular momentum \(l\) with parity \((-1)^l\): \(\lvert J_i - J_f\rvert\le l\le J_i + J_f\) and \(\pi_i = \pi_f(-1)^l\). A \(0^+\) parent can therefore reach only natural-parity states \(0^+,1^-,2^+,3^-,\ldots\) with \(l = J_f\). A nonzero \(l\) adds a centrifugal barrier \(\hbar^2l(l+1)/2mr^2\) that hinders the decay.
Example 5.2 For \({}^{238}_{\ 92}\mathrm{U}\to{}^{234}_{\ 90}\mathrm{Th}+\alpha\) (\(Q_\alpha = 4.27\) MeV, \(r_0 = 1.2\) fm), find the barrier height, the turning point and the barrier width.
Solution.\(R = 1.2(234^{1/3} + 4^{1/3})\) fm \(= 1.2(6.162 + 1.587)\) fm \(= 9.30\) fm, so \(V_C = 90\times2\times1.44/9.30\) MeV \(= 27.9\) MeV. The turning point is \(b = 259.2/4.27\) fm \(= 60.7\) fm, and the barrier is \(51.4\) fm wide.
Example 5.3\({}^{210}\mathrm{Po}\) and \({}^{212}\mathrm{Po}\) have \(Q_\alpha = 5.41\) and \(8.95\) MeV. Estimate the ratio of their half-lives.
Solution. With \(Z_d = 82\) for both, only the leading term of \(2G\) differs: \(t_{210}/t_{212}\approx\exp[3.96\times82(5.41^{-1/2} - 8.95^{-1/2})] = e^{31.0}\approx3\times10^{13}\), against the measured \(4\times10^{13}\).
5.3 Beta decay: processes and energetics
(a) The three processes
Beta decay moves a nucleus towards the line of stability at constant \(A\): \[ \begin{aligned} \beta^-:&\quad {}^{A}_{Z}\mathrm{X}\to{}^{A}_{Z+1}\mathrm{Y} + e^- + \bar\nu_e\qquad(n\to p + e^- + \bar\nu_e),\\ \beta^+:&\quad {}^{A}_{Z}\mathrm{X}\to{}^{A}_{Z-1}\mathrm{Y} + e^+ + \nu_e\qquad(p\to n + e^+ + \nu_e),\\ \text{EC}:&\quad {}^{A}_{Z}\mathrm{X} + e^-\to{}^{A}_{Z-1}\mathrm{Y} + \nu_e\qquad(p + e^-\to n + \nu_e). \end{aligned} \]
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