Appendix I: Answers to Selected Exercises

This appendix provides answers to selected exercises from each chapter. For each chapter, approximately every third or fourth problem is answered. Numerical answers include units and significant figures consistent with the given data. Brief solution sketches are included for non-trivial problems. For computational problems, expected output values are given.

Constants used throughout: $m_p = 938.272$ MeV/$c^2$, $m_n = 939.565$ MeV/$c^2$, $m_\alpha = 3727.379$ MeV/$c^2$, $m_e = 0.511$ MeV/$c^2$, $\hbar c = 197.3$ MeV$\cdot$fm, $e^2/(4\pi\epsilon_0) = 1.440$ MeV$\cdot$fm, $1\,\text{u} = 931.494$ MeV/$c^2$, $r_0 = 1.21$ fm (unless otherwise specified).

Chapter 1: The Discovery of the Nucleus

Problem 1.1. (a) Distance of closest approach: $d_0 = kz_1z_2e^2 / T = 1.44 \times 2 \times 13 / 7.7 = 4.87$ fm. (b) $R_{\text{Al}} = 1.21 \times 27^{1/3} = 1.21 \times 3.00 = 3.63$ fm. Since $d_0 = 4.87$ fm $> R = 3.63$ fm, the alpha particle does not reach the nuclear surface and Rutherford scattering is valid. (c) $b = (d_0/2)\cot(\theta/2) = (4.87/2)\cot(45^\circ) = 2.43$ fm.

Problem 1.3. (a) For 5.5 MeV alphas on gold: $a = kz_1z_2e^2/(4E_{\text{cm}}) = 1.44 \times 2 \times 79 / (4 \times 5.41) = 10.52$ fm (using $E_{\text{cm}} = 5.5 \times 197/201 = 5.41$ MeV). $d\sigma/d\Omega = a^2 / (4\sin^4(\theta/2))$: At $\theta = 45^\circ$: $\sin^4(22.5^\circ) = 0.02145$, so $d\sigma/d\Omega = 110.7^2 / (4 \times 0.02145) = 1.29 \times 10^5$ fm$^2$/sr $= 12.9$ b/sr. At $\theta = 90^\circ$: $\sin^4(45^\circ) = 0.250$, so $d\sigma/d\Omega = 1107$ fm$^2$/sr $= 1.11$ b/sr. At $\theta = 135^\circ$: $\sin^4(67.5^\circ) = 0.729$, so $d\sigma/d\Omega = 380$ fm$^2$/sr $= 0.380$ b/sr.

Problem 1.7. (b) From $T_p = 4m_n M_p T_n / (m_n + M_p)^2$ and $T_N = 4m_n M_N T_n / (m_n + M_N)^2$, dividing: $T_p/T_N = M_p(m_n + M_N)^2 / [M_N(m_n + M_p)^2]$. With $T_p = 5.7$ MeV, $T_N = 1.4$ MeV, $M_p = 1$ u, $M_N = 14$ u: solving gives $m_n = 1.15$ u. The modern value is 1.00866 u; the discrepancy arises from using approximate recoil energies.

Problem 1.13. $R = 1.21 \times A^{1/3}$ fm: (a) $^4$He: $R = 1.21 \times 1.587 = 1.92$ fm. (b) $^{27}$Al: $R = 1.21 \times 3.00 = 3.63$ fm. (c) $^{56}$Fe: $R = 1.21 \times 3.826 = 4.63$ fm. (d) $^{208}$Pb: $R = 1.21 \times 5.925 = 7.17$ fm. (e) $^{238}$U: $R = 1.21 \times 6.198 = 7.50$ fm.

Problem 1.17. Using $B = [Zm_H + Nm_n - M(A,Z)] \times 931.494$ MeV: (a) $^2$H: $B = [1.007825 + 1.008665 - 2.014102] \times 931.494 = 2.224$ MeV; $B/A = 1.112$ MeV. (b) $^3$He: $B = [2(1.007825) + 1.008665 - 3.016029] \times 931.494 = 7.718$ MeV; $B/A = 2.573$ MeV. (d) $^{16}$O: $B = [8(1.007825) + 8(1.008665) - 15.994915] \times 931.494 = 127.62$ MeV; $B/A = 7.976$ MeV. (e) $^{62}$Ni: $B = [28(1.007825) + 34(1.008665) - 61.928345] \times 931.494 = 545.26$ MeV; $B/A = 8.795$ MeV.

Problem 1.20. $S_n = [M(A-1,Z) + m_n - M(A,Z)]c^2$: (a) $S_n(^{16}\text{O}) = [15.003066 + 1.008665 - 15.994915] \times 931.494 = 15.66$ MeV. (b) $S_n(^{17}\text{O}) = [15.994915 + 1.008665 - 16.999132] \times 931.494 = 4.14$ MeV. The jump from 4.14 MeV to 15.66 MeV reveals the $N = 8$ magic number: adding the 9th neutron (beyond the closed shell) costs much less energy than removing a neutron from the closed shell.

Problem 1.23. (a) Number of fissions/s: $P / E_f = 10^9 / (200 \times 10^6 \times 1.602 \times 10^{-19}) = 3.12 \times 10^{19}$ fissions/s. (b) Mass of $^{235}$U consumed per day: $(3.12 \times 10^{19} \times 86400 \times 235) / (6.022 \times 10^{23}) = 1.05$ kg/day. (c) Coal: $10^9 / (30 \times 10^6) = 33.3$ kg/s $= 2880$ tonnes/day. The ratio is $2.88 \times 10^6 / 1.05 \approx 2.7 \times 10^6$.

Chapter 2: Nuclear Properties

Problem 1. $R_{1/2} = r_0 A^{1/3}$ and $R_{\text{rms}} = R_{1/2}\sqrt{3/5 + 7\pi^2 a^2/(3R_{1/2}^2)}$: (a) $^{12}$C: $R_{1/2} = 1.21 \times 2.289 = 2.77$ fm; $R_{\text{rms}} \approx 2.46$ fm (exp: 2.470 fm). (b) $^{56}$Fe: $R_{1/2} = 4.63$ fm; $R_{\text{rms}} \approx 3.74$ fm (exp: 3.738 fm). Agreement is within 1--2%.

Problem 4. (a) $\Delta r_{np} = R_n - R_p = 5.727 - 5.501 = 0.226 \pm 0.071$ fm. (b) Volume of skin shell: $(4\pi/3)(R_n^3 - R_p^3) = (4\pi/3)(187.8 - 166.5) = 89.0$ fm$^3$. Total neutron volume: $(126/208)(4\pi/3)(7.17)^3 = 930$ fm$^3$. Fraction $\approx 89/930 \approx 9.6\%$.

Problem 8. Using $B = Z\Delta_H + N\Delta_n - \Delta$: $^{62}$Ni: $B = 28(7289.0) + 34(8071.3) - (-66746.1) = 204092 + 274424 + 66746 = 545262$ keV $= 545.26$ MeV; $B/A = 8.795$ MeV. $^{56}$Fe: $B = 26(7289.0) + 30(8071.3) - (-60605.5) = 189514 + 242139 + 60606 = 492259$ keV; $B/A = 8.790$ MeV. $^{62}$Ni has the highest $B/A$, not $^{56}$Fe.

Problem 14. (a) $J^\pi = 1/2^-$. The odd proton fills $1p_{1/2}$ ($\ell = 1$, $j = 1/2$). Experimental: $1/2^-$. Correct. (b) $J^\pi = 3/2^+$. 19th proton fills $1d_{3/2}$ ($\ell = 2$, $j = 3/2$). Experimental: $3/2^+$. Correct. (c) $J^\pi = 9/2^+$. 41st proton fills $1g_{9/2}$. Experimental: $9/2^+$. Correct. (d) $J^\pi = 1/2^-$. Odd neutron hole in $1p_{1/2}$. Experimental: $1/2^-$. Correct.

Problem 18. (a) Schmidt value for $^3$He (neutron in $1s_{1/2}$, $j = l + 1/2$ with $l = 0$): $\mu = (0)(0) + (1/2)(-3.826) = -1.913\,\mu_N$. Experimental: $-2.128\,\mu_N$. Ratio: 1.11. (b) Schmidt for $^7$Li (proton in $1p_{3/2}$, $j = l + 1/2$ with $l = 1$): $\mu = (1)(1) + (1/2)(5.586) = 3.793\,\mu_N$. Experimental: $3.256\,\mu_N$. Ratio: 0.86.

Problem 24. (c) Isoscalar: $\mu(^3\text{H}) + \mu(^3\text{He}) = 2.979 + (-2.128) = 0.851\,\mu_N$. Free-nucleon isoscalar: $(\mu_p + \mu_n)/2 = (2.793 - 1.913)/2 = 0.440\,\mu_N$. Ratio: $0.851/0.440 = 1.93$.

Chapter 3: The Nuclear Force

Problem 3.1. (a) $R = 1.2 \times 208^{1/3} = 1.2 \times 5.925 = 7.11$ fm. (b) $E_C = (3/5)(82)(81)(1.44)/(7.11) = (3/5)(9564.5) = 806$ MeV. (c) Fraction: $806/1636.4 = 0.493$. Nearly half the binding energy must compensate Coulomb repulsion.

Problem 3.7. (a) $K = \sqrt{2\mu(V_0 - B_d)}/\hbar = \sqrt{2(469.5)(35 - 2.225)} / 197.3 = \sqrt{30730}/197.3 = 0.888$ fm$^{-1}$. $\kappa = \sqrt{2\mu B_d}/\hbar = \sqrt{2(469.5)(2.225)}/197.3 = 0.2316$ fm$^{-1}$. (c) $P_{\text{outside}} \approx 0.73$. About 73% of the deuteron wavefunction extends beyond the potential well, confirming the deuteron is a very loosely bound system.

Problem 3.9. (b) $P_D \approx 0.040$ (approximately 4% D-state admixture). This is an estimate because meson exchange currents and relativistic corrections also contribute to $\Delta\mu$.

Problem 3.14. (a) $\lambda_C = \hbar/(mc) = 197.3/m$ (with $m$ in MeV): $\pi$: $197.3/138 = 1.43$ fm (long-range attraction, $r > 1.4$ fm). $\sigma$: $197.3/475 = 0.415$ fm (medium-range attraction, $r \sim 0.4$--$1.4$ fm). $\rho$: $197.3/775 = 0.255$ fm (short-range). $\omega$: $197.3/783 = 0.252$ fm (short-range repulsion).

Problem 3.17. (a) Underbinding: $8.482 - 7.62 = 0.862$ MeV, or $0.862/8.482 = 10.2\%$. (b) $^4$He prediction with 3NF: 28.34 MeV vs. experimental 28.296 MeV, a discrepancy of only 0.044 MeV (0.16%). This is a true prediction since $^4$He was not fitted, making it a stringent test.

Chapter 4: The Semi-Empirical Mass Formula

Problem 4.1. For $^{40}$Ca: $B_{\text{SEMF}} = 630.0 - 208.2 - 78.9 + 0 + 1.8 = 344.7$ MeV. Experimental: 342.052 MeV. Error: 0.8%.

Problem 4.4. For $^{56}$Fe ($Z = 26$, $A = 56$): Volume: $15.75$ MeV/nucleon. Surface: $-17.80 \times 56^{-1/3} = -4.65$ MeV/nucleon (29.5% of $a_V$). Coulomb: $-0.711 \times 25 \times 26/56^{4/3} = -2.67$ MeV/nucleon (17.0%). Asymmetry: $-23.7 \times 16/(56^2) = -0.121$ MeV/nucleon (0.8%). Pairing: $+11.2/56^{3/2} = +0.027$ MeV/nucleon. The surface correction is the largest.

Problem 4.9. (a) $E_C = 0.711 \times 82 \times 81 / 208^{1/3} = 0.711 \times 6642 / 5.925 = 796.5$ MeV. (b) Fraction: $796.5/1636.4 = 48.7\%$. (c) Without Coulomb: $B = 1636.4 + 796.5 = 2432.9$ MeV; $B/A = 11.70$ MeV. (d) No; the most stable isobar would have $Z = A/2 = 104$ (symmetric nuclear matter).

Problem 4.12. (b) $Z_{\text{stable}} = A / [2 + (a_C/(4a_{\text{sym}}))A^{2/3}]$: $A = 20$: $Z = 20/[2 + 0.00749 \times 7.37] = 20/2.055 = 9.73 \approx 10$ (Ne). Correct. $A = 200$: $Z = 200/[2 + 0.00749 \times 34.20] = 200/2.256 = 88.6 \approx 80$ (Hg). Close.

Problem 4.18. (a) $x = Z^2/(50.88 A)$: $^{238}$U: $x = 92^2 / (50.88 \times 238) = 8464/12109 = 0.699$. $^{294}$Og: $x = 118^2/(50.88 \times 294) = 13924/14959 = 0.931$. $Z = 130$, $A = 330$: $x = 16900/16790 = 1.007$. Instantaneous fission predicted.

Chapter 5: Quantum Mechanics Review

Exercise 5.1. (a) $1s_{1/2}$: $n = 1$, $l = 0$, $j = 1/2$, $\pi = +$, degeneracy $= 2$. (d) $1f_{7/2}$: $n = 1$, $l = 3$, $j = 7/2$, $\pi = -$, degeneracy $= 8$. (f) $1g_{9/2}$: $n = 1$, $l = 4$, $j = 9/2$, $\pi = +$, degeneracy $= 10$.

Exercise 5.5. (a) $^3$He: odd neutron in $1s_{1/2}$, $J^\pi = 1/2^+$. Correct. (b) $^{15}$N: odd proton hole in $1p_{1/2}$, $J^\pi = 1/2^-$. Correct. (c) $^{17}$O: odd neutron in $1d_{5/2}$, $J^\pi = 5/2^+$. Correct. (d) $^{41}$Ca: odd neutron in $1f_{7/2}$, $J^\pi = 7/2^-$. Correct.

Exercise 5.14. (a) $2^+ \to 0^+$: $\Delta J = 2$, $\Delta\pi = +1$ (no change). Lowest multipole: $E2$. (b) $3^- \to 2^+$: $\Delta J = 1$, $\Delta\pi = -1$. Lowest: $E1$. (e) $0^+ \to 0^+$: $\Delta J = 0$, forbidden by single-photon emission since the photon carries at least $J = 1$.

Exercise 5.22. (b) $\kappa = \sqrt{2 \times 938.3 \times (30 - 5)} / 197.3 = \sqrt{46915}/197.3 = 1.098$ fm$^{-1}$. $T = \exp(-2 \times 1.098 \times 5) = e^{-10.98} = 1.70 \times 10^{-5}$. (c) At $E = 10$ MeV: $\kappa = 0.979$ fm$^{-1}$, $T = e^{-9.79} = 5.6 \times 10^{-5}$. A factor of 3.3 increase for doubling the energy.

Exercise 5.25. (a) $V_C = 1.44 / 1.2 = 1.20$ MeV $= 1200$ keV. (b) $\langle E \rangle = (3/2)(8.617 \times 10^{-5})(1.5 \times 10^7) = 1.94$ keV. The thermal energy is $\sim 600$ times smaller than the Coulomb barrier, yet fusion proceeds via quantum tunneling at the Gamow peak.

Chapter 6: The Nuclear Shell Model

Problem 1. (a) $^{40}$Ca: $Z = 20$ (magic), $N = 20$ (magic). Doubly magic. (d) $^{132}$Sn: $Z = 50$ (magic), $N = 82$ (magic). Doubly magic. (f) $^{48}$Ca: $Z = 20$ (magic), $N = 28$ (magic). Doubly magic. (h) $^{208}$Pb: $Z = 82$ (magic), $N = 126$ (magic). Doubly magic. (j) $^{56}$Ni: $Z = 28$ (magic), $N = 28$ (magic). Doubly magic.

Problem 3. (c) Cumulative: $N = 0$: 2; $N = 1$: 8; $N = 2$: 20; $N = 3$: 40; $N = 4$: 70; $N = 5$: 112; $N = 6$: 168. (d) The $N = 3$ closure at 40 does not match experiment because spin-orbit coupling splits the $1f$ level, pulling $1f_{7/2}$ down into the lower shell to create the magic number 28 instead.

Problem 7. $1g_{7/2}$: 8; $2d_{5/2}$: 6; $2d_{3/2}$: 4; $3s_{1/2}$: 2; $1h_{11/2}$: 12. Total: $8 + 6 + 4 + 2 + 12 = 32 = 82 - 50$. Confirmed.

Problem 9. (b) $^{13}$C ($N = 7$): odd neutron in $1p_{1/2}$, $J^\pi = 1/2^-$. Experimental: $1/2^-$. Correct. (d) $^{29}$Si ($N = 15$): odd neutron in $2s_{1/2}$, $J^\pi = 1/2^+$. Experimental: $1/2^+$. Correct. (f) $^{39}$K ($Z = 19$): odd proton hole in $1d_{3/2}$, $J^\pi = 3/2^+$. Experimental: $3/2^+$. Correct. (h) $^{87}$Sr ($N = 49$): odd neutron in $2p_{1/2}$ (just below $N = 50$), $J^\pi = 9/2^+$ (the 49th neutron fills $1g_{9/2}$). Experimental: $9/2^+$. Correct.

Problem 16. (a) Proton in $1d_{5/2}$ ($j = l + 1/2$, $l = 2$): $\mu = (j - 1/2)g_l + (1/2)g_s = 2(1) + (1/2)(5.586) = 4.793\,\mu_N$. (c) Neutron in $1f_{7/2}$ ($j = l + 1/2$, $l = 3$): $\mu = 3(0) + (1/2)(-3.826) = -1.913\,\mu_N$. (e) Proton in $1g_{9/2}$ ($j = l + 1/2$, $l = 4$): $\mu = 4(1) + (1/2)(5.586) = 6.793\,\mu_N$.

Chapter 7: Beyond the Single Particle

Exercise 7.5 (Calcium binding energies). (a) $S_n(N = 21) = B(^{41}\text{Ca}) - B(^{40}\text{Ca}) = 350.415 - 342.052 = 8.363$ MeV. $S_n(N = 22) = 361.895 - 350.415 = 11.480$ MeV. $S_n(N = 28) = 415.991 - 406.043 = 9.948$ MeV. The odd-even staggering and the large value at $N = 28$ (magic) are clearly visible.

Exercise 7.9. (d) Eigenvalues: $E_\pm = (E_1 + E_2)/2 \pm \sqrt{[(E_2 - E_1)/2]^2 + V_{12}^2}$. With $V_{12} = 0.5$ MeV, $\Delta E = 2.0$ MeV: $E_- = -1.06$ MeV (shift of 0.06 MeV), $E_+ = 1.06$ MeV. Mixing amplitude $|c_2|^2 = 0.015$ (1.5% admixture).

Exercise 7.13. (b) For $^{99m}$Tc, the $M4$ Weisskopf estimate gives $t_{1/2}^W \approx 5 \times 10^{-4}$ s. Experimental: 6.01 hours $= 2.16 \times 10^4$ s. Hindrance factor: $F_W = 2.16 \times 10^4 / 5 \times 10^{-4} \approx 4.3 \times 10^7$.

Exercise 7.21. (b) $\mathcal{J}_{\text{exp}} = 6\hbar^2 / (2 \times 0.0798) = 37.6\,\hbar^2$/MeV. (a) $\mathcal{J}_{\text{rigid}} = (2/5)(939)(168)(1.2 \times 168^{1/3})^2 = 87.8\,\hbar^2$/MeV. (c) Ratio: $37.6/87.8 = 0.43$. The nuclear moment of inertia is less than half the rigid-body value, a signature of nuclear superfluidity due to pairing.

Chapter 8: Collective Motion

Problem 2. (a) $E(0^+_2)/E(2^+_1) = 1224/618 = 1.98$; $E(2^+_2)/E(2^+_1) = 1312/618 = 2.12$; $E(4^+_1)/E(2^+_1) = 1416/618 = 2.29$. Harmonic vibrator predicts all three ratios equal to 2.0. The triplet is spread but centered near $2\hbar\omega$.

Problem 9. (a) Rigid rotor: $E(I) = E(2^+) \times I(I+1)/6$. $E(4^+) = 100.8 \times 20/6 = 336$ keV; $E(6^+) = 100.8 \times 42/6 = 706$ keV; $E(8^+) = 100.8 \times 72/6 = 1210$ keV; $E(10^+) = 100.8 \times 110/6 = 1848$ keV. (c) $\mathcal{J} = 6\hbar^2 / (2 \times 0.1008) = 29.8\,\hbar^2$/MeV.

Problem 16. $R_{4/2}$ values: $^{114}$Cd: $1210/558 = 2.17$ (vibrational). $^{152}$Sm: $367/122 = 3.01$ (transitional). $^{186}$W: $396/122 = 3.25$ (near-rotational). $^{178}$Hf: $307/93 = 3.30$ (rotational). $^{196}$Pt: $877/356 = 2.46$ (vibrational/transitional).

Problem 24. (a) $N_B(^{154}\text{Sm}) = (62 - 50)/2 + (92 - 82)/2 = 6 + 5 = 11$. (b) $N_B(^{168}\text{Er}) = (68 - 50)/2 + (100 - 82)/2 = 9 + 9 = 18$. (c) $^{168}$Er has the largest $N_B$ and is mid-shell for both protons and neutrons, placing it closest to the SU(3) rotational limit.

Chapter 9: Electromagnetic Transitions

Problem 1. (a) $2^+ \to 0^+$: $E2$. (b) $3^- \to 2^+$: $E1$. (c) $5/2^+ \to 1/2^-$: $E2$ (lowest: $\Delta J = 2$, parity change). (d) $4^+ \to 2^+$: $E2$. (e) $7/2^- \to 5/2^-$: $M1$. (f) $0^+ \to 0^+$: forbidden for single-photon emission. (g) $1^+ \to 0^+$: $M1$. (h) $8^+ \to 2^+$: $E6$.

Problem 8. (a) $T_W(E1)$ at $E_\gamma = 1$ MeV, $A = 50$: $T_W \approx 1.0 \times 10^{14}$ s$^{-1}$; $t_{1/2} \approx 7 \times 10^{-15}$ s. (b) $T_W(E2)$ at $E_\gamma = 1$ MeV, $A = 150$: $T_W \approx 7.3 \times 10^{7}$ s$^{-1}$; $t_{1/2} \approx 9.5 \times 10^{-9}$ s. (c) $T_W(M1)$ at $E_\gamma = 0.5$ MeV: $T_W \approx 1.8 \times 10^{10}$ s$^{-1}$; $t_{1/2} \approx 3.9 \times 10^{-11}$ s.

Problem 15. (a) Conversion electron kinetic energies: $T_K = 121.8 - 46.83 = 74.97$ keV; $T_{L_I} = 121.8 - 7.74 = 114.06$ keV; $T_{L_{II}} = 114.49$ keV; $T_{L_{III}} = 115.08$ keV. (b) $\alpha_{\text{total}} = 0.65 + 0.085 + 0.055 + 0.038 = 0.828$. Fraction as gamma ray: $1/(1 + 0.828) = 0.547$ (54.7%).

Chapter 10: Exotic Nuclei

10.7. (a) $\kappa = \sqrt{2 \times 857 \times 0.502}/197.3 = \sqrt{860.2}/197.3 = 0.149$ fm$^{-1}$; $1/\kappa = 6.73$ fm. (b) Well-bound: $\kappa = \sqrt{2 \times 900 \times 8}/197.3 = 0.608$ fm$^{-1}$; $1/\kappa = 1.64$ fm. (c) Ratio: $6.73/1.64 = 4.1$. The halo wavefunction extends $\sim 4$ times farther, explaining the anomalous size.

10.12. (a) Good magic number: $^{48}$Ca ($N = 28$, $E(2^+) = 3832$ keV), $^{40}$Ca ($N = 20$, $E(2^+) = 3904$ keV). Magic number collapsed: $^{42}$Si ($N = 28$, $E(2^+) = 770$ keV), $^{32}$Mg ($N = 20$, $E(2^+) = 885$ keV). (b) $E(2^+)$ ratio: $3832/770 = 4.98$. The factor of 5 reduction signals strong collectivity in $^{42}$Si.

10.19. (a) $V_C = 1.44 \times 70 / 7.5 = 13.4$ MeV. (b) $V_C / Q_p = 13.4 / 1.233 = 10.9$. The barrier exceeds the available energy by a factor of $\sim 11$.

Chapter 11: Superheavy Elements

Exercise 11.4. (a) $x$ values: $^{208}$Pb: $82^2/(50.88 \times 208) = 6724/10583 = 0.635$. $^{238}$U: $92^2/12109 = 0.699$. $^{252}$Cf: $98^2/12822 = 0.749$. $^{298}$Fl: $114^2/15162 = 0.857$. (b) $B_f^{\text{LD}}$: $^{208}$Pb: $0.38 \times 17.8 \times 208^{2/3} \times (1-0.635)^3 = 0.38 \times 627 \times 0.0487 = 11.6$ MeV. $^{298}$Fl: $0.38 \times 17.8 \times 298^{2/3} \times (1-0.857)^3 = 0.38 \times 795 \times 0.00293 = 0.88$ MeV. Shell correction of 6 MeV gives total $B_f \approx 6.9$ MeV.

Exercise 11.7. (a) $Q_\alpha = E_\alpha (1 + 4/A_d)$: $^{294}$Og: $Q = 11.65 \times (1 + 4/290) = 11.65 \times 1.0138 = 11.81$ MeV. $^{290}$Lv: $Q = 10.84 \times (1 + 4/286) = 10.84 \times 1.0140 = 10.99$ MeV. $^{286}$Fl: $Q = 10.19 \times (1 + 4/282) = 10.19 \times 1.0142 = 10.33$ MeV.

Exercise 11.12. (a) $E^*/8 = 35/8 \approx 4.4$; dominant channel is 4n evaporation. (b) $P_{\text{surv}}^{\text{total}} = 0.3^4 = 0.0081$. (c) $\sigma_{\text{SHE}} = 30 \times 10^{-3} \times 5 \times 10^{-4} \times 0.0081 = 1.2 \times 10^{-7}$ b $= 0.12$ nb $\approx 120$ pb.

Chapter 12: Radioactivity Fundamentals

Problem 12.1 (representative). For $^{137}$Cs ($t_{1/2} = 30.17$ yr): $\lambda = \ln 2 / t_{1/2} = 0.02297$ yr$^{-1} = 7.28 \times 10^{-10}$ s$^{-1}$. Specific activity of 1 g: $A = \lambda N = (7.28 \times 10^{-10})(6.022 \times 10^{23}/137) = 3.20 \times 10^{12}$ Bq $= 3.20$ TBq $= 86.5$ Ci.

Problem 12.4 (representative). Secular equilibrium in the $^{238}$U chain: In secular equilibrium, all daughter activities equal the parent activity. $A(^{238}\text{U}) = A(^{234}\text{Th}) = A(^{234}\text{Pa}) = \ldots$. For 1 kg of natural uranium ore in secular equilibrium: $A = 1.24 \times 10^4$ Bq per isotope in the chain.

Problem 12.7 (representative). $^{14}$C dating: Modern specific activity $A_0 = 0.226$ Bq/g carbon. A sample measures $A = 0.141$ Bq/g. Age: $t = (t_{1/2}/\ln 2)\ln(A_0/A) = (5730/0.6931)\ln(0.226/0.141) = 8267 \times 0.472 = 3900$ years.

Chapter 13: Alpha Decay

Exercise 13.2. (a) $^{210}$Po: $Q = [209.982874 - 205.974465 - 4.002603] \times 931.494 = 0.005806 \times 931.494 = 5.407$ MeV. $T_\alpha = Q \times 206/210 = 5.304$ MeV. $T_d = 0.103$ MeV. (b) $^{238}$U: $Q = [238.050788 - 234.043601 - 4.002603] \times 931.494 = 0.004584 \times 931.494 = 4.270$ MeV. $T_\alpha = 4.198$ MeV. (d) $^{208}$Pb: $Q = [207.976652 - 203.973494 - 4.002603] \times 931.494 = 0.000555 \times 931.494 = 0.517$ MeV. Energetically allowed but with an extremely long half-life ($>10^{20}$ years).

Exercise 13.3. (b) $T_\alpha = 3.271 \times 144/148 = 3.183$ MeV. $T_d = 3.271 - 3.183 = 0.088$ MeV. (c) Setting $T_d = Q \times 4/A = 0.1$ MeV with $Q = 5$ MeV: $4/A = 0.02$, so $A = 200$.

Chapter 14: Beta Decay

Problem 14.1 (representative). (a) For $^{14}$C $\to$ $^{14}$N + $e^- + \bar{\nu}_e$: $Q = [14.003242 - 14.003074] \times 931.494 = 0.156$ MeV. Classification: $0^+ \to 1^+$, $\Delta J = 1$, no parity change. Gamow-Teller allowed transition. $\log ft = 9.04$ (superallowed GT).

Problem 14.4 (representative). For $0^+ \to 0^+$ superallowed Fermi decay of $^{14}$O: $ft = 3072 \pm 2$ s. From $ft = K / (G_V^2 |M_F|^2)$ with $|M_F|^2 = 2$ for $T = 1 \to T = 1$ transitions: $G_V / (\hbar c)^3 = 1.136 \times 10^{-5}$ GeV$^{-2}$.

Chapter 15: Gamma Decay and Internal Conversion

Problem 15.1 (representative). The $11/2^-$ isomer in $^{137}$Ba at 661.66 keV decays to the $3/2^+$ ground state. $\Delta J = 4$, parity change. Lowest multipole: $E4$ (but $M4$ with $\Delta J = 4$ also allowed; the dominant transition is $M4$ since the actual shell model matrix element favors it for this configuration).

Problem 15.4 (representative). Weisskopf estimate for $E2$ at 1.33 MeV in $A = 60$ ($^{60}$Ni): $T_W(E2) = 7.3 \times 10^7 \times 60^{4/3} \times 1.33^5 = 7.3 \times 10^7 \times 229 \times 4.18 = 7.0 \times 10^{10}$ s$^{-1}$; $t_{1/2} = 9.9 \times 10^{-12}$ s $\approx 10$ ps.

Chapter 16: Radiation Interactions with Matter

Problem 16.1 (representative). Mean free path of 1 MeV photons in lead ($\mu/\rho = 0.0708$ cm$^2$/g, $\rho = 11.35$ g/cm$^3$): $\lambda = 1/(\mu) = 1/(0.0708 \times 11.35) = 1.24$ cm. The half-value layer is $\lambda \ln 2 = 0.86$ cm.

Problem 16.4 (representative). Bragg peak for 200 MeV protons in water: Range $\approx 26$ cm. The dose rate at the Bragg peak is approximately 3--4 times the entrance dose, illustrating why proton therapy is advantageous for deep-seated tumors.

Chapter 17: Nuclear Reaction Fundamentals

Problem 17.1 (representative). For $^{12}$C$(d,p)^{13}$C with $T_d = 10$ MeV: $Q = 2.722$ MeV. $E_{\text{cm}} = 10 \times 12/14 = 8.571$ MeV. Since $Q > 0$, no threshold.

Problem 17.4 (representative). Threshold energy for $^{12}$C$(\gamma,n)^{11}$C: $Q = -(B(^{12}\text{C}) - B(^{11}\text{C})) = -S_n(^{12}\text{C}) = -18.72$ MeV. Threshold $E_\gamma = |Q|(1 + |Q|/(2M_T c^2)) = 18.72 \times (1 + 18.72/11178) = 18.75$ MeV.

Chapter 18: Compound Nucleus Reactions

Problem 18.1 (representative). Breit-Wigner cross section at resonance: $\sigma_{\text{res}} = \pi \lambda^2 g (\Gamma_a \Gamma_b / \Gamma^2)$. For a neutron resonance in $^{238}$U at $E_0 = 6.67$ eV with $\Gamma_n = 1.5$ meV, $\Gamma_\gamma = 26$ meV, $J = 1/2$, $I = 0$: $g = (2J+1)/((2I+1)(2s+1)) = 2/2 = 1$. $\lambda = \hbar c / \sqrt{2m_n E_0} = 197.3 / \sqrt{2 \times 939.565 \times 6.67 \times 10^{-6}} = 55700$ fm. $\sigma_0 = \pi (55700)^2 \times 1 \times (1.5 \times 26)/27.5^2 = 1.70 \times 10^5$ b.

Chapter 19: Direct Reactions

Problem 19.1 (representative). $(d,p)$ stripping to the $1d_{5/2}$ state: $\ell = 2$ transfer. Angular distribution peaks at $\theta_{\text{cm}} = 0$ and shows a diffraction pattern characteristic of $\ell = 2$: the first minimum is near $\theta \approx 20^\circ$ for typical kinematics. The spectroscopic factor extracted from the cross section normalization is $C^2S \approx 0.5$--$0.9$ for single-particle states near closed shells.

Chapter 20: Nuclear Fission

Problem 20.1. (a) $^{208}$Pb: $x = 82^2/(50.88 \times 208) = 0.635$. Has a barrier. (b) $^{235}$U: $x = 92^2/(50.88 \times 235) = 0.708$. Has a barrier. (c) $^{252}$Cf: $x = 98^2/(50.88 \times 252) = 0.749$. Has a barrier, but spontaneous fission competes. (d) $^{298}$Fl: $x = 0.857$. Liquid drop barrier $\approx 1$ MeV; exists only due to shell correction.

Problem 20.3. (c) For $x = 0.7$: coefficient $(2E_S - E_C)/5 = E_S(2 - 2x)/5 = E_S \times 0.6/5 > 0$. The quadrupole deformation energy cost decreases as $x \to 1$, reducing the barrier.

Problem 20.7 (representative). Energy released per fission of $^{235}$U: Kinetic energy of fragments $\approx 170$ MeV, prompt neutrons $\approx 5$ MeV, prompt gammas $\approx 7$ MeV, beta particles $\approx 8$ MeV, neutrinos $\approx 12$ MeV, delayed gammas $\approx 7$ MeV. Recoverable energy $\approx 200$ MeV (neutrino energy escapes).

Chapter 21: Nuclear Fusion

Problem 21.1 (representative). $Q$ for D-T fusion: $Q = B(^4\text{He}) - B(^2\text{H}) - B(^3\text{H}) = 28.296 - 2.225 - 8.482 = 17.59$ MeV. The neutron carries $E_n = Q \times 4/5 = 14.07$ MeV and the alpha carries $E_\alpha = 3.52$ MeV.

Problem 21.4 (representative). Gamow peak for D-T at $T = 10$ keV ($\approx 10^8$ K): $E_G = 64$ keV, width $\Delta E_G = 38$ keV. The astrophysical S-factor is $S(E_G) \approx 11$ MeV$\cdot$b.

Chapter 22: Stellar Nucleosynthesis

Problem 22.1 (representative). PP-I net: $4p \to ^4\text{He} + 2e^+ + 2\nu_e$. $Q = 4(938.272) - 3727.379 - 2(0.511) = 3753.088 - 3728.401 = 24.69$ MeV. Including positron annihilation ($2 \times 2 \times 0.511 = 2.044$ MeV): $Q_{\text{total}} = 26.73$ MeV.

Problem 22.5 (representative). Triple-alpha rate: The Hoyle state resonance at $E_r = 0.3795$ MeV above the $3\alpha$ threshold controls the rate. At $T_9 = 0.1$ ($10^8$ K), the rate is negligibly small; at $T_9 = 1$, the exponential Boltzmann factor $\exp(-E_r/k_BT) = \exp(-4.41) = 0.012$ allows significant production.

Chapter 23: Explosive Nucleosynthesis

Problem 23.1 (representative). In the r-process, the waiting-point approximation at $N = 82$ gives an accumulation of material near $A = 130$. If $S_n \approx 2$ MeV defines the path: $\beta$-decay half-lives of $\sim 100$ ms set the timescale. The abundance peak width depends on the spread of $\beta$-decay rates, typically $A = 128$--$134$.

Problem 23.4 (representative). GW170817 ejecta mass: $\sim 0.05\,M_\odot$ of r-process material. Total mass of gold produced: $\sim 10^{-5}\,M_\odot \approx 3 \times 10^{25}$ kg, roughly 10 Earth masses of gold.

Chapter 24: Big Bang Nucleosynthesis

Problem 24.1 (representative). Primordial $^4$He mass fraction: With $n/p = 1/7$ at freeze-out, all neutrons go into $^4$He. $Y_p = 2(n/p)/(1 + n/p) = 2/7 / (8/7) = 2/8 = 0.25$. Observed: $Y_p = 0.245 \pm 0.003$.

Problem 24.4 (representative). Deuterium bottleneck: BBN cannot proceed until $T$ drops below $\sim 0.07$ MeV ($8 \times 10^8$ K) because photo-disintegration ($\gamma + d \to p + n$) destroys deuterium at higher temperatures. This occurs at $t \approx 3$ minutes after the Big Bang.

Chapter 25: Neutron Stars

Problem 25.1 (representative). Average density of a $1.4\,M_\odot$ neutron star with $R = 12$ km: $\rho = M / [(4/3)\pi R^3] = 2.8 \times 10^{30} / [(4/3)\pi (1.2 \times 10^4)^3] = 3.9 \times 10^{17}$ kg/m$^3$. This is $\rho/\rho_0 = 3.9 \times 10^{17} / 2.3 \times 10^{17} = 1.7$ times nuclear saturation density.

Problem 25.4 (representative). TOV maximum mass: For a polytropic EOS $P = K\rho^\gamma$ with $\gamma = 2$ (stiff): $M_{\text{max}} \approx 2.2\,M_\odot$. With $\gamma = 1.5$ (soft): $M_{\text{max}} \approx 1.6\,M_\odot$. The observed 2.0 $M_\odot$ pulsars rule out the softest equations of state.

Chapter 26: Nuclear Energy

Problem 26.1. (a) $\eta = \nu \sigma_f / (\sigma_f + \sigma_c) \times (\text{fuel fraction})$. At 20% enrichment: $\eta = 2.43 \times 584 / (584 + 99) = 2.43 \times 0.855 = 2.08$. At 3.5%: $\eta = 2.43 \times 584 / (584 + 99 + (96.5/3.5) \times 2.68) = 2.43 \times 584 / 757 = 1.87$.

Problem 26.3. (a) $N_{238} = N_C/400 = 8.03 \times 10^{22}/400 = 2.01 \times 10^{20}$ cm$^{-3}$. $p = \exp[-(2.01 \times 10^{20} \times 277 \times 10^{-24})/(0.158 \times 0.385)] = \exp[-0.915] = 0.401$. (c) $k_\infty = \eta f \varepsilon p = 1.34 \times 0.90 \times 1.03 \times 0.401 = 0.498$. Not critical.

Chapter 27: Nuclear Medicine

Problem 27.1 (representative). $^{18}$F-FDG half-life: 109.8 min. A patient receives 370 MBq at 8:00 AM. Activity at scan time (9:30 AM, $t = 90$ min): $A = 370 \times 2^{-90/109.8} = 370 \times 0.567 = 210$ MBq.

Problem 27.4 (representative). Proton Bragg peak: A 230 MeV proton beam has a range of $\sim 33$ cm in water. The Bragg peak width (80%--80%) is approximately 1.5 cm. The peak-to-entrance dose ratio is $\sim 3.5:1$.

Chapter 28: Nuclear Security

Problem 28.1 (representative). Critical mass of $^{235}$U (bare sphere): $M_c = (\pi/k_{\infty})^2 \times (4\pi D / (3\Sigma_a))$ where $D = 1/(3\Sigma_{tr})$. For weapons-grade uranium ($>90\%$): $M_c \approx 52$ kg. With a beryllium reflector, this reduces to $\sim 15$ kg.

Problem 28.4 (representative). Isotopic fingerprint: $^{235}$U/$^{238}$U ratio in reactor-grade uranium $= 3$--$5\%$; weapons-grade $> 90\%$. The $^{234}$U/$^{235}$U ratio varies by production method and serves as a forensic signature.

Chapter 29: Radiation in the Environment

Problem 29.1 (representative). Annual dose from $^{40}$K in the body: Body contains $\sim 140$ g of potassium, 0.0117% is $^{40}$K. Activity: $A = 4.4 \times 10^3$ Bq. Average energy per decay: $\sim 0.5$ MeV. Annual dose: $\sim 0.17$ mSv.

Problem 29.4 (representative). Radon in a basement: If $^{222}$Rn concentration is 150 Bq/m$^3$ (EPA action level = 148 Bq/m$^3$), a person breathing 0.5 m$^3$/hr for 7000 hr/yr inhales $5.25 \times 10^5$ Bq$\cdot$hr, contributing $\sim 3$ mSv/yr to lung dose.

Chapter 30: Accelerators and Experimental Techniques

Problem 30.1 (representative). FRIB beam rigidity: For $^{238}$U$^{92+}$ at 200 MeV/nucleon, $\gamma = 1 + T/(m_p c^2) = 1 + 200/931.5 = 1.215$, $\beta = 0.566$, $B\rho = \gamma m v / (qe) = 1.215 \times 238 \times 931.5 \times 0.566 / (92 \times 299.8) = 5.53$ T$\cdot$m.

Problem 30.4 (representative). Penning trap mass resolution: For a cyclotron frequency $\nu_c = 1$ MHz and measurement time $T = 1$ s, the resolving power is $R = \nu_c T = 10^6$. For $A = 100$, this gives $\delta m / m = 10^{-6}$, or $\delta m \approx 0.1$ keV. Modern Penning traps achieve $\delta m < 1$ keV for isotopes with $t_{1/2} > 100$ ms.

Chapter 31: The Standard Model and Nuclear Physics

Problem 31.2. (a) Current quark mass in proton: $2m_u + m_d = 2(2.2) + 4.7 = 9.1$ MeV. Fraction: $9.1/938.3 = 0.97\%$. Over 99% of the proton mass comes from QCD binding energy (gluon field energy and quark kinetic energy), not from the Higgs mechanism.

Problem 31.5 (representative). Running coupling: $\alpha_s(M_Z) = 0.118$. At $Q = 1$ GeV (nuclear scale): $\alpha_s \approx 0.5$. At $Q = 100$ GeV: $\alpha_s \approx 0.10$. The growth of $\alpha_s$ at low $Q$ is responsible for confinement and explains why perturbative QCD cannot be directly applied to nuclear structure.

Chapter 32: Fundamental Symmetries

Problem 32.1 (representative). Wu experiment: In the $^{60}$Co $\beta$-decay experiment, the angular distribution of electrons relative to the nuclear spin axis was $W(\theta) = 1 + \alpha (v/c) \cos\theta$ with $\alpha \approx -1$ for the $5^+ \to 4^+$ GT transition, demonstrating maximal parity violation.

Problem 32.4 (representative). Neutrinoless $\beta\beta$: For $^{76}$Ge with $Q_{\beta\beta} = 2039$ keV, the signal is a monoenergetic peak at $E = Q_{\beta\beta}$. Current limits: $t_{1/2} > 1.8 \times 10^{26}$ years (GERDA), corresponding to $\langle m_{\beta\beta} \rangle < 0.08$--$0.18$ eV depending on nuclear matrix element.

Chapter 33: Frontiers of Nuclear Physics

Problem 33.1 (representative). Open questions: (1) What is the equation of state of dense matter? Constrained by $M_{\text{max}} > 2.0\,M_\odot$ and $R_{1.4} \approx 12$ km from NICER. (2) Where are the neutron drip lines for $Z > 10$? FRIB will extend measurements to $Z \approx 40$. (3) Does the island of stability exist? Predicted near $Z = 114$, $N = 184$; current experiments reach $N = 177$.

Chapter 34: Capstone Project

No numerical answers; this is a portfolio project. Expected outputs include: a complete level scheme for the chosen nucleus, SEMF comparison, decay chain simulation, and connection to at least one application or astrophysical process.

Chapter 35: Reading the Literature

Problem 35.1 (representative). NNDC NuDat query: Looking up $^{132}$Sn yields $J^\pi = 0^+$, $t_{1/2} = 39.7$ s, $S_n = 7.31$ MeV, $E(2^+_1) = 4041$ keV, $B(E2; 0^+ \to 2^+) = 220$ e$^2$fm$^4$, confirming doubly-magic character.

Problem 35.4 (representative). The AME2020 mass table lists 3557 masses (2457 experimental, 1100 estimated). The ENSDF contains evaluated level schemes for $\sim 3300$ nuclides. The ENDF/B-VIII.0 library provides neutron cross sections for $\sim 560$ materials.