Your Answer : (Not Answered)

Correct Answer : B

The spins are \( \frac{3}{2} \) for \( ^{11}_{5}\text{B} \) and \( \frac{5}{2} \) for \( ^{27}_{13}\text{Al} \).

In the shell model, paired nucleons contribute nothing, so an odd-A nucleus takes the spin of its single unpaired nucleon.

• \( ^{11}\text{B} \): 5 protons and 6 neutrons. The neutrons fill 1s_1/2 (2) and 1p_3/2 (4) completely. The protons put 2 in 1s_1/2 and 3 in 1p_3/2, so the odd proton is in 1p_3/2: spin \( \frac{3}{2} \).
• \( ^{27}\text{Al} \): 13 protons and 14 neutrons. The neutrons fill up to 1d_5/2 exactly (2 + 4 + 2 + 6). The protons put 8 in the first three levels and 5 in 1d_5/2, so the odd proton is in 1d_5/2: spin \( \frac{5}{2} \).

The other options give spins of levels the odd proton does not occupy, or, like \( \frac{11}{2} \), values not available in these shells.

Your Answer : (Not Answered)

Correct Answer : D

The reaction fails on strangeness and the third component of isospin.

Check each quantity, initial against final, for \( \pi^{+} + n \rightarrow K^{0} + p \):

• Charge: \( +1 + 0 = 0 + 1 \), conserved.
• Baryon number: \( 0 + 1 = 0 + 1 \), conserved.
• Strangeness: \( 0 + 0 \ne +1 + 0 \), violated.
• Isospin \( I_{3} \): \( +1 - \frac{1}{2} = +\frac{1}{2} \) against \( -\frac{1}{2} + \frac{1}{2} = 0 \), violated.

Since charge and baryon number are both conserved, every option naming either of them is wrong. A strong reaction must conserve both strangeness and I₃, so it cannot occur.

Your Answer : (Not Answered)

Correct Answer : B

The strangeness of the \( \Omega^{-} \) is −3.

Strangeness is conserved in this strong interaction. The proton and the pions have \( S = 0 \), the \( K^{-} \) has \( S = -1 \), and the \( K^{0} \) and \( K^{+} \) each have \( S = +1 \):

\[ \begin{aligned} 0 + (-1) &= S_{\Omega} + 1 + 1 + 0 + 0 \\ S_{\Omega} &= -3 \end{aligned} \]

This agrees with the quark content of the \( \Omega^{-} \), sss, three strange quarks. +3 has the wrong sign, and ±2 would follow from leaving out one of the two kaons.

Your Answer : (Not Answered)

Correct Answer : B

The decay is a second forbidden Gamow-Teller transition.

The spin changes by ΔJ = 3 (from 3 to 0, the paper's O⁺ being 0⁺), and the parity does not change (+ to +). The classification is:

• Allowed: ΔJ = 0, ±1, no parity change (Fermi for ΔJ = 0 with no spin flip, Gamow-Teller for ΔJ = 0, ±1).
• First forbidden: parity changes, ΔJ up to 2.
• Second forbidden: no parity change, ΔJ = 2 or 3.

ΔJ = 3 with no parity change needs two units of orbital angular momentum plus the electron and neutrino spins aligned, which is a second forbidden (unique) Gamow-Teller decay. An allowed Fermi or Gamow-Teller decay cannot change J by 3, and a first forbidden decay requires a parity change.

Your Answer : (Not Answered)

Correct Answer : A

The barrier height is about 26 MeV.

The barrier is the Coulomb energy of the alpha particle (charge 2e) touching the daughter nucleus (charge 82e). With \( e^{2}/4\pi\epsilon_{0} = 1.44 \) MeV fm and \( r_{0} \approx 1.2 \) fm:

\[ \begin{aligned} R &= 1.2\,(206^{1/3} + 4^{1/3}) \approx 9.0\ \text{fm} \\ V &= \frac{1.44 \times 2 \times 82}{9.0} \approx 26\ \text{MeV} \end{aligned} \]

The alpha particles of Po-210 have only about 5·3 MeV, far below this, which is why alpha decay proceeds by quantum tunnelling. 40, 42 and 80 MeV would need a much smaller contact radius.

Your Answer : (Not Answered)

Correct Answer : B

The time is closest to 6000 years.

Radioactive decay follows \(N = N_{0}e^{-\lambda t}\). Decaying to 25% means \(N/N_{0} = 1/4\):

\[ \begin{aligned} t &= \frac{\ln 4}{\lambda} = \frac{1.386}{2.31 \times 10^{-4}} \\ &\approx 6000\ \text{years} \end{aligned} \]

Equivalently, the half-life is \(0.693/\lambda = 3000\) years, and falling to a quarter takes two half-lives. 3000 is one half-life, the time to 50%. The initial number of atoms does not matter.

Your Answer : (Not Answered)

Correct Answer : A

The required circuit is (A): \(D_{1}\) AND (NOT \(D_{2}\)).

'Scattered at angle \(\theta\) only' means \(D_{1}\) fires and \(D_{2}\) does not. The output must be 1 exactly when \(D_{1} = 1\) and \(D_{2} = 0\), which is \(D_{1}\cdot\overline{D_{2}}\): invert \(D_{2}\) and AND it with \(D_{1}\).

(B) gives \(\overline{D_{1}}D_{2}\), which selects the unscattered particles going straight to \(D_{2}\). (C) is a NOR, \(\overline{D_{1} + D_{2}}\), which is 1 only when neither detector fires. (D), \(D_{1} + \overline{D_{2}}\), is also 1 when nothing is detected at all.

Your Answer : (Not Answered)

Correct Answer : B

The minimum photon energy is \(B\left(1 + \dfrac{B}{2mc^{2}}\right)\).

The photon must supply the binding energy B and also the recoil kinetic energy of the products, because its momentum E/c must be conserved. At threshold the two nucleons move off together with total momentum E/c and total rest energy mc² + B. Energy conservation with the bound state at rest:

\[ \begin{aligned} E + mc^{2} &= \sqrt{(mc^{2} + B)^{2} + E^{2}} \\ E &= B + \frac{B^{2}}{2mc^{2}} = B\left(1 + \frac{B}{2mc^{2}}\right) \end{aligned} \]

B alone ignores the recoil. The minus sign would make the threshold smaller than the binding energy, which energy conservation forbids, and B - mc² is negative for any real nucleus.

Your Answer : (Not Answered)

Correct Answer : B

The ratio of kinetic energies is 141/93.

The initial momentum is negligible, and so is the neutrons' momentum afterwards, so the two fragments fly apart with equal and opposite momenta of magnitude p. Non-relativistically KE = p²/2M, so with the same p the kinetic energy is inversely proportional to the mass:

\[ \begin{aligned} \frac{KE_{Rb}}{KE_{Cs}} &= \frac{M_{Cs}}{M_{Rb}} \\ &\approx \frac{141}{93} \end{aligned} \]

The lighter rubidium nucleus carries more energy. 93/141 inverts the ratio, 1 would need equal masses, and 0 would mean one fragment stays at rest, which momentum conservation forbids.

Your Answer : (Not Answered)

Correct Answer : C

The nuclear force is independent of the nucleons' charge.

Mirror nuclei, such as \(^{3}\)H and \(^{3}\)He or \(^{13}\)C and \(^{13}\)N, have their proton and neutron numbers interchanged. Once the Coulomb energy is allowed for, their energy levels are nearly identical, which shows that the nn and pp interactions are the same: the strong force does not distinguish a proton from a neutron. This is charge symmetry, part of the wider idea of charge independence.

Swapping protons and neutrons leaves spins and parities of the nucleons unchanged, so mirror nuclei say nothing about dependence on spin or parity; and the tiny proton-neutron mass difference is not what the comparison tests.

Your Answer : (Not Answered)

Correct Answer : C

The first four magic numbers are 2, 8, 14, 20.

In a cubical box the levels are \(E \propto n_{x}^{2} + n_{y}^{2} + n_{z}^{2}\) with each \(n \ge 1\). Count the distinct arrangements (degeneracy), and double for the two spin states:

• (1,1,1), sum 3: 1 state, 2 nucleons, total 2.
• (2,1,1) and permutations, sum 6: 3 states, 6 nucleons, total 8.
• (2,2,1), sum 9: 3 states, 6 nucleons, total 14.
• (3,1,1), sum 11: 3 states, 6 nucleons, total 20.

Shell closures, the magic numbers in this model, are therefore 2, 8, 14, 20. The sequence 2, 8, 20, 28 is the observed set from the real nuclear shell model with spin-orbit coupling, which a simple box does not reproduce.

Your Answer : (Not Answered)

Correct Answer : B

The correct option is \(X_{1} \to Y_{1}\) is a Fermi transition and \(X_{2} \to Y_{2}\) is a Gamow-Teller transition.

In allowed beta decay there is no change of parity, and the selection rules depend on how the electron and neutrino spins combine. In a Fermi transition their spins are antiparallel (total 0), so \(\Delta J = 0\). In a Gamow-Teller transition they are parallel (total 1), so \(\Delta J = 0, \pm 1\), except that \(0 \to 0\) is forbidden.

\(0^{+} \to 0^{+}\) (\(X_{1} \to Y_{1}\)) cannot be Gamow-Teller, so it is a pure Fermi transition. \(0^{+} \to 1^{+}\) (\(X_{2} \to Y_{2}\)) has \(\Delta J = 1\), which a Fermi transition cannot provide, so it is pure Gamow-Teller.

That rules out every other combination. (This is a multiple-select question; the key accepts this option alone.)

Your Answer : (Not Answered)

Correct Answer : C

The correct statement is \(2^{+} \rightarrow 0^{+}\) is an electric quadrupole transition.

A photon of multipolarity \(L\) must satisfy \(|J_{i} - J_{f}| \le L \le J_{i} + J_{f}\) (and \(L \ge 1\)). An electric \(EL\) transition changes parity by \((-1)^{L}\) and a magnetic \(ML\) transition by \((-1)^{L+1}\).

For \(2^{+} \rightarrow 0^{+}\), only \(L = 2\) is allowed, and there is no parity change, so it must be E2, electric quadrupole. M2 would need a parity change.

For \(4^{+} \rightarrow 2^{+}\), \(L\) runs from 2 to 6 with no parity change, so the allowed multipoles are E2, M3, E4, M5, E6, and E2 dominates. An electric octupole (E3) or a magnetic quadrupole (M2) would both change parity, so the first two statements are wrong, as is the last.

Your Answer : (Not Answered)

Correct Answer : A

The allowed process is \(\pi^{-} + p \to \pi^{0} + n\).

This charge exchange reaction conserves charge (−1 + 1 = 0 + 0), baryon number (1 = 1), and isospin and strangeness, so it proceeds by the strong interaction and is observed.

\(\pi^{0} \to 3\gamma\) is forbidden by charge conjugation: the \(\pi^{0}\) has C = +1, and each photon carries C = −1, so three photons have C = −1. \(p + \bar{p} \to \Lambda^{0} + \Lambda^{0}\) has baryon number 0 before and 2 after (and strangeness 0 before, −2 after). \(p + \bar{p} \to \gamma\) cannot conserve both energy and momentum: in the centre of mass frame the total momentum is zero, but a single photon always carries momentum.

Your Answer : (Not Answered)

Correct Answer : A

The baryon with \(S = -1\) is \(\Sigma^{*0}\).

Strangeness counts strange quarks, each contributing \(S = -1\). The \(\Sigma^{*0}\) (the spin-3/2 partner of the \(\Sigma^{0}\)) has quark content \(uds\), one strange quark, so \(S = -1\).

The neutron (\(udd\)) and the \(\Delta^{0}\) (\(udd\)) contain no strange quark and have \(S = 0\). The \(\Xi^{*0}\) (\(uss\)) has two strange quarks and \(S = -2\).

Your Answer : (Not Answered)

Correct Answer : A

The possible process is \(\nu_{e} + {}_{z}P^{A} \rightarrow {}_{z+1}Q^{A} + e^{-}\).

This is neutrino capture, the inverse of beta decay: inside the nucleus a neutron absorbs the electron neutrino and turns into a proton, emitting an electron (ν_e + n → p + e^-). Charge is conserved (0 + z = z + 1 - 1), the mass number is unchanged, and the lepton number (+1 on both sides) and electron flavour are conserved.

Producing a positron with a neutrino would violate lepton number: the left side has +1 and the right side -1. Creating an e⁺e^- pair leaves the neutrino's lepton number unaccounted for, and the last option destroys the neutrino and creates only a photon, again violating lepton number.

Your Answer : (Not Answered)

Correct Answer : A

The true statement is p is odd under parity and μ is odd under time reversal; it is the only one the key accepts in this multiple-select question.

The electric dipole moment \(\mathbf{p} = \sum q\mathbf{r}\) is built from position vectors. Parity sends \(\mathbf{r} \rightarrow -\mathbf{r}\), so \(\mathbf{p}\) is odd under parity; time reversal leaves positions unchanged, so \(\mathbf{p}\) is even under T.

The magnetic dipole moment comes from circulating currents or spin, \(\boldsymbol{\mu} \propto \mathbf{r} \times \mathbf{v}\) or \(\propto \mathbf{S}\), an axial vector like angular momentum. It is even under parity (both factors change sign) but odd under time reversal, since reversing time reverses every velocity and spin.

So 'μ is even under time reversal' and 'p is even under parity' are both false, which rules out the other three statements.

Your Answer : (Not Answered)

Correct Answer : B

The flavour wavefunction is \(\frac{1}{\sqrt{3}}(uss + sus + ssu)\).

Strangeness -2 means two strange quarks; with total isospin projection +1/2 the third quark is a \(u\). The baryon decuplet is the totally symmetric flavour multiplet of SU(3) (its spin part is symmetric too, and colour supplies the antisymmetry), so the state must be the symmetric sum of all distinct orderings of \(u, s, s\), normalised by \(1/\sqrt{3}\).

The two \(1/\sqrt{2}\) options leave out the \(sus\) ordering, so they are neither symmetric nor complete; \(uss - ssu\) is also antisymmetric under exchange of the first and third quarks. The option with \(-sus\) is not symmetric under exchanging neighbouring quarks.

Your Answer : (Not Answered)

Correct Answer : D

\(X\) is the muon antineutrino, \(\bar{\nu}_{\mu}\).

The decay is \(\mu^{+} \to e^{+} + \nu_{e} + \bar{\nu}_{\mu}\). Check the lepton family numbers. On the left, \(\mu^{+}\) has muon number \(L_{\mu} = -1\) and electron number \(L_{e} = 0\). On the right, \(e^{+}\) has \(L_{e} = -1\) and \(\nu_{e}\) has \(L_{e} = +1\), cancelling, so the remaining particle must carry \(L_{\mu} = -1\) and no charge: the muon antineutrino.

A photon carries no lepton number, so \(\gamma\) leaves \(L_{\mu}\) unbalanced. \(\bar{\nu}_{e}\) would make \(L_{e} = -1\) and leave \(L_{\mu}\) unbalanced. \(\nu_{\mu}\) has \(L_{\mu} = +1\), the wrong sign; it is the partner in the \(\mu^{-}\) decay.

Your Answer : (Not Answered)

Correct Answer : C

\(\Xi^{0}\) is at point R.

In this drawing strangeness increases upward and the third component of isospin, \(I_{3}\), to the right. The top row, with \(\Delta^{++}\) at the right end, is the four \(\Delta\) baryons with \(S = 0\). Going down, the rows are the three \(\Sigma^{*}\) (\(S = -1\)), the two \(\Xi^{*}\) (\(S = -2\)) and the single \(\Omega^{-}\) (\(S = -3\)) at the bottom.

The \(\Xi\) row holds P and R. The two members have \(I_{3} = -\tfrac{1}{2}\) (\(\Xi^{-}\)) and \(I_{3} = +\tfrac{1}{2}\) (\(\Xi^{0}\)); charge \(Q = I_{3} + (B + S)/2\) gives \(Q = 0\) for the right-hand one, so \(\Xi^{0}\) is R and \(\Xi^{-}\) is P.

Q is the \(\Sigma^{*0}\) in the middle of the second row, and S is the \(\Omega^{-}\).