Theory, worked examples and 18 previous-year questions with full solutions — IIT JAM, GATE, JEST, TIFR and CSIR NET-JRF, 2020 onwards.
Theory
No exact theory of the nucleus exists, so nuclear physics proceeds through models, each simple enough to be worked out and judged by how many measured properties it explains. In the independent-particle picture (Fermi gas, shell model) each nucleon moves in the average potential of all the others; this explains nucleon kinetic energies, the magic numbers, and the spins, parities and moments of ground states. In the collective picture (liquid drop, collective model) the nucleus behaves as a whole; this explains binding-energy systematics, \(\beta\) stability, fission, and rotational and vibrational spectra. The Pauli principle reconciles the two: deep inside the Fermi sea a nucleon cannot scatter, because the states into which it would scatter are already occupied.
3.1 The Fermi gas model
(a) Fermi momentum and Fermi energy
Protons and neutrons are treated as two independent gases of non-interacting fermions confined to the nuclear volume, each filling its own flat-bottomed well, two particles (spin up and down) per spatial state, up to the Fermi energy \(E_F\). The proton well is shallower because of the Coulomb repulsion, which also produces the barrier outside the nucleus (Figure 3.1). In a stable nucleus the two Fermi levels coincide, about \(B'\approx8\) MeV below zero; otherwise a neutron would convert into a proton by \(\beta^-\) decay or vice versa by \(\beta^+\) decay or electron capture. This explains at once why heavy nuclei need more neutrons than protons.
Figure 3.1 The Fermi gas model: neutrons and protons fill separate wells up to a common Fermi level. The proton well is shallower because of the Coulomb repulsion; \(B'\) is the binding energy of the last nucleon.
For a box of volume \(V\) the allowed momenta form a lattice with one spatial state per volume \((2\pi\hbar)^3/V\) of momentum space. Counting the states inside the sphere \(p\le p_F\), with two spin states, gives \(n = Vp_F^3/(3\pi^2\hbar^3)\) particles of one kind, so \[\boxed{\,p_F = \hbar\left(3\pi^2\,\frac{n}{V}\right)^{1/3},\qquad E_F = \frac{\hbar^2}{2m}\left(3\pi^2\,\frac{n}{V}\right)^{2/3}\,}\] Both depend only on the number density. With \(V = \tfrac43\pi r_0^3A\) and \(n = Z\) or \(N\), \(p_F^{(p)} = (\hbar/r_0)(9\pi Z/4A)^{1/3}\) and \(p_F^{(n)} = (\hbar/r_0)(9\pi N/4A)^{1/3}\); for \(N = Z = A/2\)\[\boxed{\,p_F = \frac{\hbar}{r_0}\left(\frac{9\pi}{8}\right)^{1/3}\approx250\ \text{MeV}/c,\qquad E_F = \frac{\hbar^2}{2mr_0^2}\left(\frac{9\pi}{8}\right)^{2/3}\approx33\ \text{MeV}\propto\frac{1}{r_0^2}\,}\] with \(k_F = p_F/\hbar\approx1.3\ \text{fm}^{-1}\) and \(v_F/c = p_Fc/mc^2\approx0.27\), so nucleons are only mildly relativistic and \(v = p/m\) suffices. The mass number has cancelled: because nuclear density is constant (saturation), \(p_F\) and \(E_F\) are the same for all nuclei. The depth of the well is \(V_0 = E_F + B'\approx40\) MeV. In a heavy nucleus the neutrons, being more numerous, have the higher Fermi energy (for \({}^{208}\text{Pb}\) about 38 MeV against 28 MeV); the Coulomb energy raises the proton well by the difference.
(b) Total energy, asymmetry energy, and the single-particle estimate
Since the number of states between \(p\) and \(p+dp\) is proportional to \(p^2dp\), the mean kinetic energy is \(\langle E\rangle = \int_0^{p_F}(p^2/2m)p^2dp\big/\int_0^{p_F}p^2dp\), that is, \[\boxed{\,\langle E\rangle = \tfrac35E_F\approx20\ \text{MeV},\qquad E_{\text{tot}} = \tfrac35\,nE_F\,}\] If \(V\propto n\), \(E_F\) is fixed and the total energy is proportional to \(n\) (doubling \(n\) doubles \(E_{\text{tot}}\)); this is the origin of the volume term below. At fixed volume, by contrast, \(E_F\propto n^{2/3}\) and \(E_{\text{tot}}\propto n^{5/3}\).
Writing \(N,Z = \tfrac{A}{2}(1\pm x)\) with \(x = (N-Z)/A\) and \(E_F^{(n,p)} = E_0(1\pm x)^{2/3}\), where \(E_0\) is the value for \(N = Z\), the total kinetic energy \(\tfrac35E_0\tfrac{A}{2}[(1+x)^{5/3} + (1-x)^{5/3}]\) expands to \[E_{\text{kin}}\approx\tfrac35E_0A + \tfrac13E_0\,\frac{(N-Z)^2}{A}.\] A neutron excess therefore costs energy proportional to \((N-Z)^2/A\) with coefficient \(E_0/3\approx11\) MeV, about half the empirical asymmetry coefficient.
The Fermi gas must not be confused with the single-particle uncertainty estimate. A nucleon confined to a region of radius \(R = r_0A^{1/3}\) has \(p\sim\hbar/R\propto A^{-1/3}\) and hence a kinetic energy (mean or root-mean-square) \[\boxed{\,E\sim\frac{\hbar^2}{2mR^2}\propto A^{-2/3}\,}\] This scaling is what questions on “a nucleon confined in a nucleus of mass number \(A\)” intend. The Fermi gas adds the Pauli principle, which raises the typical momentum to \(p_F\), several times larger and independent of \(A\).
Example 3.1 In nuclear matter \(k_F = 1.33\ \text{fm}^{-1}\). Find the Fermi speed, the Fermi energy, the mean kinetic energy and the total nucleon density (\(\hbar c = 197.3\) MeV fm, \(mc^2 = 939\) MeV).
Solution.\(p_Fc = \hbar ck_F = 262\) MeV, so \(v_F/c = p_Fc/mc^2 = 0.28\) (the relativistic value, \(262/975 = 0.27\), is hardly different). \(E_F = (p_Fc)^2/2mc^2 = 36.7\) MeV and \(\langle E\rangle = \tfrac35E_F = 22\) MeV. For one kind of nucleon \(n/V = k_F^3/3\pi^2 = 0.080\ \text{fm}^{-3}\); with both kinds the density is \(0.16\ \text{fm}^{-3}\).
3.2 The liquid-drop model and the semi-empirical mass formula
The constant nuclear density and the nearly constant \(B/A\approx8\) MeV both reflect the short range and saturation of the nuclear force, as in a drop of incompressible liquid. Weizsäcker’s formula builds \(B(A,Z)\) from terms with a clear physical origin. The volume term\(a_VA\) expresses saturation (an unsaturated force would give \(\propto A^2\)). Nucleons at the surface have fewer neighbours, so the surface term\(-a_SA^{2/3}\) is proportional to the area \(4\pi R^2\propto A^{2/3}\); per nucleon it is \(a_SA^{-1/3}\), largest for light nuclei. The Coulomb term follows from the energy of a uniformly charged sphere, \(U = \tfrac35q^2/(4\pi\varepsilon_0R)\), with \(Z^2\to Z(Z-1)\) since a proton does not repel itself: \[B_C = -a_C\,\frac{Z(Z-1)}{A^{1/3}},\qquad a_C = \frac35\,\frac{e^2}{4\pi\varepsilon_0r_0} = \frac35\times\frac{1.44\ \text{MeV fm}}{1.2\ \text{fm}} = 0.72\ \text{MeV};\] many questions write \(a_CZ^2/A^{1/3}\). The asymmetry term\(-a_A(A-2Z)^2/A\) comes from the Pauli principle (Section 3.1(b)) and from the extra strength of the \(np\) interaction; it vanishes for \(N = Z\) and is symmetric in \(N\leftrightarrow Z\). The pairing term reflects the extra binding of like nucleons paired to \(J = 0\): even–even nuclei are the most bound and odd–odd the least (of about 250 stable nuclides only \({}^2\text{H}\), \({}^6\text{Li}\), \({}^{10}\text{B}\), \({}^{14}\text{N}\) are odd–odd). Altogether \[\boxed{\,B(A,Z) = a_VA - a_SA^{2/3} - a_C\,\frac{Z(Z-1)}{A^{1/3}} - a_A\,\frac{(A-2Z)^2}{A} + \delta,\qquad \delta = \begin{cases}+a_PA^{-1/2} & \text{e–e}\\ 0 & \text{odd }A\\ -a_PA^{-1/2} & \text{o–o}\end{cases}}\] and \(M(A,Z)c^2 = ZM({}^1\text{H})c^2 + (A-Z)m_nc^2 - B\). Questions that supply their own coefficients or forms are worked with those.
Term
Form
Coefficient (MeV)
Origin
Volume
\(+a_VA\)
15.8
saturated short-range attraction
Surface
\(-a_SA^{2/3}\)
18.3
fewer neighbours at the surface
Coulomb
\(-a_CZ(Z-1)A^{-1/3}\)
0.714
proton–proton repulsion
Asymmetry
\(-a_A(A-2Z)^2A^{-1}\)
23.2
Pauli principle
Pairing
\(\pm a_PA^{-1/2}\) (or \(\pm a_PA^{-3/4}\))
12 (or 33.5)
pairing of like nucleons
The older Fermi form of the pairing term, \(\pm33.5A^{-3/4}\) MeV, is still common in questions. The surface term causes the low \(B/A\) of light nuclei, the Coulomb term (per nucleon \(\propto Z^2/A^{4/3}\propto A^{2/3}\)) the decline beyond iron. The formula fits measured binding energies to about 1 per cent for \(A>20\), except near the magic numbers.
3.3 Mass parabolas and \(\beta\) stability
(a) The most stable isobar
At fixed \(A\), collecting powers of \(Z\) in \(M(A,Z)c^2\) gives a parabola, \[\boxed{\,M(A,Z)c^2 = \alpha A + \beta Z + \gamma Z^2 - \delta,\qquad \gamma = \frac{4a_A}{A} + \frac{a_C}{A^{1/3}},\quad -\beta = 4a_A + \frac{a_C}{A^{1/3}} + (m_n - M_{\text{H}})c^2\,}\] whose minimum \(Z_0 = -\beta/2\gamma\) marks the most stable isobar. Dropping the small terms (or, equivalently, setting \(\partial B/\partial Z = 0\) at fixed \(A\) with the Coulomb term \(a_CZ^2/A^{1/3}\)), \[\boxed{\,Z_0\approx\frac{A}{2 + \dfrac{a_C}{2a_A}\,A^{2/3}}\approx\frac{A}{2 + 0.0154\,A^{2/3}}\,}\] so \(Z_0\approx A/2\) for light nuclei and \(Z_0/A\approx0.39\) for \(A = 238\). Whatever form a question gives (with \((N-Z)^2\), a ratio \(a_A/a_C\), or \(Z(Z-1)\)), the method is the same: differentiate \(B\) with respect to \(Z\) at fixed \(A\) and set the derivative to zero; the isobar with the nearest integer \(Z\) has the largest binding.
(b) Odd and even \(A\); \(\beta\)-decay energies
For odd \(A\), \(\delta = 0\) and all isobars lie on one parabola, \(Mc^2 = M(A,Z_0)c^2 + \gamma(Z - Z_0)^2\). Isobars on the neutron-rich side decay by \(\beta^-\), those on the proton-rich side by \(\beta^+\) or electron capture, so for odd \(A\) there is only one \(\beta\)-stable isobar. With atomic masses \[\boxed{\,\begin{aligned}Q_{\beta^-}(Z\to Z+1) &= 2\gamma\left(Z_0 - Z - \tfrac12\right),\\ Q_{\text{EC}}(Z\to Z-1) &= 2\gamma\left(Z - Z_0 - \tfrac12\right),\qquad Q_{\beta^+} = Q_{\text{EC}} - 2m_ec^2\end{aligned}\,}\] If \(Z_0\) is an integer, the final \(\beta^-\) step into it releases exactly \(\gamma\) (the coefficient of \(Z^2\)), the step before it \(3\gamma\); the electron capture from \(Z_0+1\) also releases \(\gamma\).
For even \(A\), even–even and odd–odd isobars lie on two parabolas separated by \(2\lvert\delta\rvert\) (Figure 3.2), and the decays zigzag between them. Hence two (occasionally three) even–even isobars can be stable, an odd–odd nucleus near the minimum can decay both ways (\({}^{64}\text{Cu}\), \({}^{40}\text{K}\)), and stable odd–odd nuclei exist only among the lightest.
Figure 3.2 Atomic masses of the isobars with \(A = 106\). Odd–odd and even–even nuclei lie on two parabolas separated by \(2\delta\); both \({}^{106}\text{Pd}\) and \({}^{106}\text{Cd}\) (open circles) are stable.
(c) Mirror nuclei
Mirror nuclei have \(Z\) and \(N\) interchanged. The volume and surface terms depend only on \(A\), and \((A-2Z)^2 = (N-Z)^2\) is the same for both, so only the Coulomb term differs: \[\boxed{\,B(A,Z) - B(A,Z' = N) = -\frac{a_C}{A^{1/3}}\left(Z^2 - N^2\right)\,}\]
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