CSIR NET Physical Sciences Nuclear and Particle Physics Syllabus

What is the binding energy of the deuteron, and what does the absence of an excited bound state tell us?

About $2.22\,\text{MeV}$. The deuteron has only one (weakly) bound state and no excited bound states, indicating the nuclear potential well is just barely deep enough to bind one nucleon pair.

Why does the existence of the deuteron's quadrupole moment prove the nuclear force is not purely central?

A purely central force would give a pure $\ell = 0$ ($S$) state, which is spherically symmetric with zero quadrupole moment. The observed nonzero $Q$ requires a $D$-state admixture, produced by a non-central tensor force.

Why is there no bound state of the deuteron in the spin-singlet ($S = 0$) configuration?

The spin-triplet ($S = 1$) nuclear force is slightly more attractive (partly via the tensor force). The singlet potential is too shallow to support a bound state — the singlet deuteron is virtual/unbound.

What are the nuclear magic numbers, and what do they signify?

$2, 8, 20, 28, 50, 82, 126$. Nuclei with a magic number of protons and/or neutrons have closed shells, giving extra stability (closed-shell binding), analogous to noble-gas electron configurations.

List three pieces of experimental evidence for nuclear shell structure.

(1) Extra abundance/stability and high binding energy of nuclei at magic numbers; (2) discontinuities in nucleon separation energies and in neutron-capture cross sections at magic numbers; (3) doubly-magic nuclei (e.g. $\ce{^{16}O}$, $\ce{^{208}Pb}$) being exceptionally stable with high first-excited-state energies.

What crucial ingredient must be added to the shell-model potential to reproduce the higher magic numbers (28, 50, 82, 126)?

A strong spin-orbit coupling term, $\propto \vec{\ell} \cdot \vec{s}$. It splits each level into $j = \ell + \tfrac{1}{2}$ and $j = \ell - \tfrac{1}{2}$, with the higher-$j$ state lowered, creating the observed shell gaps.

In the single-particle shell model, how is the ground-state spin-parity of an odd-A nucleus predicted?

It is given by the $j^{\pi}$ of the single unpaired nucleon in the highest occupied orbital, since paired nucleons couple to $J = 0$. Parity is $(-1)^{\ell}$ of that orbital.

Using the shell model, what is the predicted ground-state spin-parity of $\ce{^{17}O}$ (8 protons, 9 neutrons)?

$J^{\pi} = \tfrac{5}{2}^{+}$. The 9th neutron occupies the $1d_{5/2}$ orbital ($j = \tfrac{5}{2}$, $\ell = 2$ so parity $+$), and protons (8, magic) and 8 neutrons are paired.

State the Schmidt-limit prediction the shell model makes for the magnetic moment of an odd-A nucleus.

The nuclear magnetic moment equals that of the single unpaired nucleon in its orbital (the Schmidt value), computed from $g_\ell$ and $g_s$ for $j = \ell \pm \tfrac{1}{2}$. Most measured moments lie between the two Schmidt lines.

Name two important successes of the single-particle shell model.

It correctly predicts the magic numbers (with spin-orbit coupling), the ground-state spins and parities of most odd-A nuclei, and the existence of isomeric (long-lived metastable) states (islands of isomerism).

State two key limitations/failures of the single-particle shell model.

(1) It fails for highly deformed (collective) nuclei far from closed shells, where rotational/vibrational behavior dominates; (2) it does not accurately predict magnetic moments (deviation from Schmidt lines), quadrupole moments, or transition rates without configuration mixing/collective corrections.

What is the energy-level formula for the rotational spectrum of an even-even deformed nucleus?

$$E_J = \frac{\hbar^{2}}{2\mathcal{I}} J(J+1)$$ where $\mathcal{I}$ is the moment of inertia. For even-even ground bands, only $J = 0, 2, 4, 6, \dots$ (positive parity) occur.

For an even-even nuclear rotational band, what is the ratio $E_{4^+}/E_{2^+}$, and what does it indicate?

$\dfrac{E_{4^+}}{E_{2^+}} = \dfrac{4\cdot 5}{2\cdot 3} = \dfrac{20}{6} \approx 3.33$. A measured ratio close to $3.33$ signals a well-deformed rotational nucleus (versus $\approx 2.0$ for a vibrator).

Write the general expression for the energy released ($Q$-value) in alpha decay and state its significance.

$Q_\alpha = \left[ M(Z,A) - M(Z-2, A-4) - M_\alpha \right] c^{2}$. Alpha decay is energetically allowed only when $Q_\alpha > 0$; the released energy is shared between the alpha particle and the recoiling daughter.

State the Geiger-Nuttall law relating alpha-decay half-life to decay energy.

$\log_{10} \lambda = a - b\, Q_\alpha^{-1/2}$ (equivalently a linear relation between $\log_{10} t_{1/2}$ and $1/\sqrt{Q_\alpha}$). Higher decay energy gives an exponentially shorter half-life.

How does Gamow's quantum theory explain alpha decay?

By quantum-mechanical tunneling: the preformed alpha particle is trapped behind the Coulomb potential barrier and has a small but finite probability per assault of tunneling through it. The exponential sensitivity of the tunneling probability to $Q_\alpha$ explains the Geiger-Nuttall law.

Write the three modes of beta decay and the particles emitted in each.

$\beta^-$: $n \to p + e^- + \bar{\nu}_e$. $\beta^+$: $p \to n + e^+ + \nu_e$. Electron capture (EC): $p + e^- \to n + \nu_e$. In each the mass number $A$ is unchanged while $Z$ changes by $\pm 1$.

What experimental puzzle led Pauli to postulate the neutrino in beta decay?

The continuous energy spectrum of the emitted beta particle (electrons emerge with a range of energies up to a maximum). To conserve energy, momentum, and angular momentum, Pauli proposed an undetected neutral particle — the (anti)neutrino — carrying the balance.

Define Fermi and Gamow-Teller transitions in beta decay by their spin change $\Delta S$.

Fermi transitions: the emitted lepton pair has total spin $S = 0$ (antiparallel), so $\Delta J = 0$ (no nuclear spin flip). Gamow-Teller transitions: lepton pair spin $S = 1$ (parallel), allowing $\Delta J = 0, \pm 1$ (but not $0 \to 0$).

State the selection rules for an allowed beta decay transition.

$\Delta \ell = 0$ (no orbital angular momentum carried off by leptons), hence no parity change ($\Delta \pi = \text{no}$). $\Delta J = 0$ (Fermi) or $\Delta J = 0, \pm 1$ excluding $0 \to 0$ (Gamow-Teller).

What characterizes a 'forbidden' beta decay, and how does its rate compare to allowed decays?

Forbidden decays carry off orbital angular momentum ($\Delta \ell \geq 1$), occurring when allowed transitions are not possible. They have much longer half-lives (smaller matrix elements). $n$-th forbidden implies $\Delta \ell = n$ and parity change $(-1)^{n}$.

What is the basic process of gamma decay, and what nuclear quantity changes?

An excited nucleus de-excites to a lower energy state by emitting a photon (gamma ray): $\ce{^{A}X^{*} -> ^{A}X + \gamma}$. Neither $Z$ nor $A$ changes — only the internal energy state. The photon energy equals the level difference (minus small recoil).

State the angular-momentum and parity selection rules for gamma (multipole) emission.

Angular momentum: $|J_i - J_f| \leq L \leq J_i + J_f$ with $L \geq 1$ (no monopole $L=0$ single-photon emission; $0 \to 0$ is forbidden). Parity: electric $EL$ requires $\Delta\pi = (-1)^{L}$; magnetic $ML$ requires $\Delta\pi = (-1)^{L+1}$.

Why is a single-photon $0^+ \to 0^+$ gamma transition strictly forbidden, and what process occurs instead?

A photon carries at least one unit of angular momentum ($L \geq 1$), so it cannot connect two $J = 0$ states. The nucleus de-excites instead by internal conversion or (if energetically allowed) electron-positron pair emission.

What is internal conversion, and how does it compete with gamma emission?

Internal conversion is an alternative de-excitation in which the nuclear excitation energy is transferred directly to an atomic (e.g. K-shell) electron, ejecting it. It competes with gamma emission and dominates for low-energy transitions, high $Z$, and especially for forbidden ($0 \to 0$) transitions.

Define the internal conversion coefficient $\alpha$.

$\alpha = \dfrac{\lambda_e}{\lambda_\gamma}$, the ratio of the internal-conversion (electron emission) rate to the gamma-emission rate. It increases with atomic number $Z$ and with higher multipolarity $L$, and decreases with transition energy.

Write the general form of a nuclear reaction and define its $Q$-value.

$a + X \to Y + b$, often written $X(a,b)Y$. The $Q$-value is $Q = (m_a + m_X - m_Y - m_b)c^{2}$. $Q > 0$ is exothermic (energy released); $Q < 0$ is endothermic and requires a threshold incident energy.

Describe the compound-nucleus mechanism (Bohr) of nuclear reactions.

The projectile is absorbed by the target to form an excited intermediate compound nucleus that lives long enough to 'forget' its formation. It then decays into outgoing particles independently of how it was formed — the two stages (formation and decay) are independent.

What is the Bohr independence hypothesis for compound-nucleus reactions?

The decay of the compound nucleus is independent of its mode of formation; the cross section factorizes as $\sigma(a,b) = \sigma_{CN}(a)\, \dfrac{\Gamma_b}{\Gamma}$, where $\sigma_{CN}(a)$ is the formation cross section and $\Gamma_b/\Gamma$ the branching ratio for decay channel $b$.

Contrast compound-nucleus reactions with direct reactions.

Compound-nucleus reactions: long-lived intermediate, energy shared among all nucleons, isotropic (forward-backward symmetric) angular distribution, sharp resonances. Direct reactions (e.g. stripping/pickup like $(d,p)$): fast surface interaction with few nucleons, strongly forward-peaked angular distribution, smooth energy dependence.