Chapter 6. Nuclear Reactions, Fission and Fusion

Theory, worked examples and 5 previous-year questions with full solutions — IIT JAM, GATE, JEST, TIFR and CSIR NET-JRF, 2020 onwards.

Theory

6.1 Nuclear reactions and conservation laws

A nuclear reaction is a process in which a projectile \(a\) strikes a target nucleus \(A\) and, through the nuclear force, produces a light particle \(b\) and a residual nucleus \(B\). It is written \(a + A\to B + b\) or compactly \(A(a,b)B\); Rutherford’s first transmutation (1919) is \({}^{14}\mathrm{N}(\alpha,p){}^{17}\mathrm{O}\). The pair \(a + A\) forms the entrance channel and each possible pair of products an exit channel; several exit channels are usually open at once, each with its own probability.

In elastic scattering, \(A(a,a)A\), the particles and their internal states are unchanged; in inelastic scattering, \(A(a,a')A^*\), the target is left excited and the kinetic energy is reduced by the excitation energy. In radiative capture, such as \((n,\gamma)\), the projectile is absorbed and \(\gamma\) rays are emitted; its inverse is photodisintegration, \((\gamma,n)\) or \((\gamma,p)\). In transfer reactions nucleons pass between projectile and target: stripping, such as \((d,p)\), when the projectile loses a nucleon, and pickup, such as \((p,d)\), when it gains one. At high energy spallation chips off many nucleons. A heavy nucleus may split into two comparable fragments (fission) and two light nuclei may combine (fusion). By mechanism, reactions are compound-nucleus reactions (slow, energy shared by all nucleons) or direct reactions (fast, involving a few surface nucleons).

Every reaction conserves electric charge (total \(Z\)), nucleon number (total \(A\), more generally baryon number), total energy including rest energy, linear momentum and total angular momentum; lepton number is conserved when leptons take part. Strong and electromagnetic reactions also conserve parity, and approximately isospin. The total angular momentum of the entrance channel is \(\mathbf J = \mathbf s_a + \mathbf J_A + \mathbf l\) and its parity \(\pi_a\pi_A(-1)^l\), and the same \(J^\pi\) must appear in the exit channel. Magnetic dipole and electric quadrupole moments are properties of individual states and are not conserved. Conservation of \(Z\) and \(A\) identifies an unknown product: in \({}^{27}_{13}\mathrm{Al}(\alpha,n)X\), \(X\) has \(Z = 15\), \(A = 30\), so it is \({}^{30}_{15}\mathrm{P}\).

6.2 The Q-value

For \(a + A\to B + b\) with the target at rest, conservation of total energy gives \((m_ac^2 + K_a) + m_Ac^2 = (m_Bc^2 + K_B) + (m_bc^2 + K_b)\). The Q-value is the excess of initial over final rest energy, equal to the gain in kinetic energy: \[ \boxed{\,Q = \big[(m_a + m_A) - (m_B + m_b)\big]c^2 = K_B + K_b - K_a\,} \] A reaction with \(Q>0\) is exoergic and can proceed even with a projectile of vanishing energy; one with \(Q<0\) is endoergic and has a threshold (Section 6.3). Elastic scattering has \(Q = 0\) and inelastic scattering to a level \(E^*\) has \(Q = -E^*\). If \(K_a\) and the energy of one product are measured, energy conservation alone gives the energy of the other: \(K_B = K_a + Q - K_b\).

Tables list atomic masses \(M\). Because charge is conserved, the electrons carried by the neutral atoms balance on the two sides provided a free proton is entered as a hydrogen atom \(M({}^1\mathrm{H})\) and an \(\alpha\) particle as \(M({}^4\mathrm{He})\); then \[ \boxed{\,Q = \big[M(a) + M(A) - M(B) - M(b)\big]c^2\,} \] with electron binding energies negligible. The same holds for \(\alpha\) decay, \(\beta^-\) decay and electron capture, but in \(\beta^+\) decay an electron mass is left over on each side, so that \(Q_{\beta^+} = [M({\rm parent}) - M({\rm daughter}) - 2m_e]c^2\), or, with nuclear masses, \(Q_{\beta^+} = (M_P - M_D - m_e)c^2\). Since the numbers of protons and of neutrons are separately conserved in strong reactions, the nucleon masses also cancel when \(M = Zm_p + Nm_n - B/c^2\) is inserted, and \[ \boxed{\,Q = \sum B_{\rm final} - \sum B_{\rm initial}\,} \] with \(B = 0\) for a free nucleon. For a heavy nucleus of \(A\) nucleons and binding energy per nucleon \(b\) splitting into fragments with \(b_1\), \(b_2\), \(Q = A_1b_1 + A_2b_2 - Ab\).

Example 6.1 Find the \(Q\)-value of \({}^{14}\mathrm{N}(n,p){}^{14}\mathrm{C}\), given \(M({}^{14}\mathrm{N}) = 14.003074\) u, \(M({}^{14}\mathrm{C}) = 14.003242\) u, \(M({}^1\mathrm{H}) = 1.007825\) u, \(m_n = 1.008665\) u.

Solution. The left side carries 7 atomic electrons and the right side \(6 + 1 = 7\), so atomic masses may be used with \(M({}^1\mathrm{H})\) for the proton: \(Q = (14.003074 + 1.008665) - (14.003242 + 1.007825) = 0.672\times10^{-3}\) u \(= 0.626\) MeV. The reaction is exoergic and proceeds with thermal neutrons; it produces \({}^{14}\mathrm{C}\) in the atmosphere.

Example 6.2 The reaction \({}^{11}\mathrm{B}(p,n){}^{11}\mathrm{C}\) is followed by \({}^{11}\mathrm{C}\to{}^{11}\mathrm{B} + e^+ + \nu_e\) with maximum positron kinetic energy 0.96 MeV. With \(m_ec^2 = 0.511\) MeV and \((m_n - m_p)c^2 = 1.293\) MeV, find \(Q\) of the \((p,n)\) reaction.

Solution. The end-point energy equals the decay \(Q\) (the neutrino takes nothing and the recoil is negligible): with nuclear masses, \((M_{\rm C} - M_{\rm B} - m_e)c^2 = 0.96\) MeV, so \((M_{\rm C} - M_{\rm B})c^2 = 1.471\) MeV. Then \(Q = (M_{\rm B} + m_p - M_{\rm C} - m_n)c^2 = -1.471 - 1.293 = -2.76\) MeV. In general a \((p,n)\) reaction whose product returns to the target by \(\beta^+\) decay has \(Q = -[T_{\max}(e^+) + m_ec^2 + (m_n - m_p)c^2]\) and is always endoergic.

6.3 Kinematics and threshold energy

(a) The Q-equation

Let \(b\) be emitted at angle \(\theta\) to the beam, the target being at rest. Momentum conservation is the triangle \(\mathbf p_a = \mathbf p_b + \mathbf p_B\), so \(p_B^2 = p_a^2 + p_b^2 - 2p_ap_b\cos\theta\). With \(p^2 = 2mK\) this eliminates the unobserved \(K_B\) from \(Q = K_B + K_b - K_a\) and gives the Q-equation \[ \boxed{\,Q = K_b\left(1 + \frac{m_b}{m_B}\right) - K_a\left(1 - \frac{m_a}{m_B}\right) - \frac{2}{m_B}\sqrt{m_am_bK_aK_b}\,\cos\theta\,} \] which is simplest at \(\theta = 90^\circ\); mass numbers may replace the masses. Particle groups of lower energy correspond to \(Q' = Q - E^*\) and measure the excited levels of \(B\). When the initial momentum is negligible (a slow projectile, or a decay or fission at rest), the two products have equal and opposite momenta and share the energy inversely as their masses: \[ \boxed{\,K_b = Q\,\frac{m_B}{m_B + m_b},\qquad K_B = Q\,\frac{m_b}{m_B + m_b}\,} \] so the lighter product takes the larger share.

(b) Centre-of-mass energy and threshold

A projectile of kinetic energy \(K_a\) striking a target at rest gives the centre of mass the velocity \(V = m_av_a/(m_a + m_A)\). Momentum conservation keeps the kinetic energy of this motion, \(K_am_a/(m_a + m_A)\), unchanged, so only the centre-of-mass energy \[ \boxed{\,K_{\rm cm} = K_a\,\frac{m_A}{m_a + m_A} = \tfrac12\mu v_a^2\,} \] is available for conversion into rest energy or excitation. An endoergic reaction becomes possible when

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Chapter 6 Nuclear Reactions, Fission and Fusion