Quiz — Chapter 15: Gamma Decay and Internal Conversion
Instructions: Select the best answer for each question. Each question has exactly one correct answer.
Q1. A nucleus in a $3^+$ excited state de-excites to a $1^+$ ground state. What is the dominant multipole character of the gamma-ray transition?
(a) E1 (b) M1 (c) E2 (d) M2
Q2. A gamma-ray transition between a $5/2^-$ state and a $1/2^+$ state involves a parity change. The lowest allowed multipole is:
(a) E1 (b) M1 (c) E2 (d) M2
Q3. A $0^+ \to 0^+$ nuclear transition:
(a) Proceeds by E0 (electric monopole) single-photon emission (b) Is strictly forbidden for single-photon emission and must proceed by internal conversion or internal pair production (c) Proceeds by M1 radiation because there is no parity change (d) Proceeds by two simultaneous E1 photons
Q4. Internal conversion is best described as:
(a) The photoelectric absorption of a gamma ray by an atomic electron (b) The direct transfer of nuclear transition energy to a bound atomic electron, without a real photon being produced (c) The Compton scattering of a gamma ray by an inner-shell electron (d) The creation of an electron-positron pair by a gamma ray in the nuclear Coulomb field
Q5. The internal conversion coefficient $\alpha$ is defined as:
(a) The probability of gamma-ray emission (b) The ratio $N_\gamma / N_e$ (gamma rays per conversion electron) (c) The ratio $N_e / N_\gamma$ (conversion electrons per gamma ray) (d) The total number of conversion electrons emitted
Q6. Which of the following conditions increases the internal conversion coefficient?
(a) Lower atomic number $Z$ (b) Lower multipole order $\lambda$ (c) Higher transition energy $E_\text{tr}$ (d) Higher atomic number $Z$
Q7. The kinetic energy of a K-shell conversion electron from a nuclear transition of energy $E_\text{tr}$ is:
(a) $T_e = E_\text{tr}$ (b) $T_e = E_\text{tr} - B_K$, where $B_K$ is the K-shell binding energy (c) $T_e = E_\text{tr} + B_K$ (d) $T_e = E_\text{tr} - 2m_e c^2$
Q8. The $K/L$ conversion coefficient ratio $\alpha_K/\alpha_L$ is useful because it:
(a) Is independent of nuclear structure and depends only on transition energy (b) Depends strongly on multipole character but weakly on transition energy, making it a good diagnostic of multipolarity (c) Is always equal to 1 for all transitions (d) Determines the nuclear spin change
Q9. A nuclear isomer is:
(a) A nucleus with the same mass number but different atomic number (b) A nucleus in a metastable excited state with a measurably long half-life (c) A nucleus with the same number of protons but different number of neutrons (d) A nucleus undergoing spontaneous fission
Q10. $^{99\text{m}}$Tc has a half-life of 6.01 hours because:
(a) It undergoes very slow beta decay (b) The isomeric transition requires high multipolarity (M4) due to the large spin difference between the isomeric state and ground state (c) It is a doubly magic nucleus (d) The transition energy is very high, making gamma emission inefficient
Q11. The 140.5 keV gamma ray from $^{99\text{m}}$Tc is ideal for SPECT imaging because:
(a) It has an energy too high for gamma cameras but is easily detected by Geiger counters (b) Its half-life is measured in microseconds, allowing rapid imaging (c) Its energy is optimal for penetrating tissue and being efficiently detected by gamma cameras, and the 6-hour half-life balances image acquisition time against patient dose (d) It is an E1 transition with the highest possible transition rate
Q12. The Mossbauer effect is:
(a) The emission of gamma rays from radioactive nuclei in free atoms (b) Recoilless nuclear resonance emission and absorption of gamma rays when nuclei are bound in a crystal lattice (c) The scattering of gamma rays by free electrons (Compton scattering) (d) The pair production of electrons and positrons by high-energy gamma rays
Q13. For a free $^{57}$Fe nucleus emitting a 14.413 keV gamma ray, the recoil energy is approximately:
(a) 14.413 keV (b) 1.96 meV (c) 4.66 neV (d) 0.097 mm/s
Q14. The recoil-free fraction $f$ in the Mossbauer effect:
(a) Is always 1.0 for any nucleus in a crystal (b) Increases with increasing gamma-ray energy (c) Increases with increasing Debye temperature and decreasing gamma-ray energy (d) Is independent of temperature
Q15. The Mossbauer effect is practical for $^{57}$Fe (14.413 keV) but not for $^{60}$Co (1332 keV) primarily because:
(a) Cobalt is not a metal (b) The recoil energy scales as $E_\gamma^2$, making the recoil-free fraction negligibly small for high-energy transitions (c) $^{60}$Co has too short a half-life (d) $^{60}$Co does not have a crystal structure
Q16. In the Pound-Rebka experiment, the gravitational redshift was measured using the Mossbauer effect. The fractional energy shift for a height difference $h$ is:
(a) $\Delta E / E = v/c$ (b) $\Delta E / E = gh/c^2$ (c) $\Delta E / E = h/R_E$ where $R_E$ is the Earth's radius (d) $\Delta E / E = mc^2/E_\gamma$
Q17. The isomer shift in Mossbauer spectroscopy is sensitive to:
(a) The nuclear spin (b) The s-electron density at the nucleus, which depends on the oxidation state and bonding of the atom (c) The applied external magnetic field (d) The temperature of the sample only
Q18. In $^{57}$Fe Mossbauer spectroscopy, a magnetically ordered sample shows a six-line pattern because:
(a) There are six isotopes of iron (b) The nuclear ground state ($I = 1/2$) splits into 2 sublevels and the excited state ($I = 3/2$) splits into 4 sublevels, with $\Delta m = 0, \pm 1$ selection rules allowing 6 transitions (c) Six gamma rays are emitted simultaneously (d) The crystal has six-fold symmetry
Q19. The energy resolution of $^{57}$Fe Mossbauer spectroscopy ($\Delta E/E \sim 10^{-13}$) exceeds that of other spectroscopic techniques because:
(a) Iron nuclei are the heaviest nuclei that exist (b) The Mossbauer effect eliminates recoil broadening, leaving only the natural linewidth, which is extremely narrow for nuclear transitions (c) Gamma-ray detectors have perfect energy resolution (d) The crystal lattice amplifies the gamma-ray energy
Q20. Which of the following is NOT a correct statement about E0 transitions?
(a) They are forbidden for single-photon emission (b) They connect states with the same spin and parity (c) They can proceed by internal conversion (d) They are the dominant de-excitation mechanism for all nuclear transitions
Answer Key
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(c) $\Delta J = |3-1| = 2$ to $3+1 = 4$; no parity change selects E2, M3, E4, ... Dominant is E2.
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(c) $\Delta J = |5/2 - 1/2| = 2$ to $5/2 + 1/2 = 3$; parity change selects E2, M3, E4. Dominant is E2. (Note: E1 requires $\lambda = 1$, excluded by $\Delta J \ge 2$; M1 also excluded and has wrong parity.)
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(b) A photon must carry $\lambda \ge 1$, but the triangle rule $|0-0| \le \lambda \le 0+0$ gives $\lambda = 0$, which is impossible for a single photon.
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(b) Internal conversion is a direct electromagnetic coupling between the nucleus and the atomic electron; no real photon is involved.
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(c) $\alpha = N_e/N_\gamma$, the number of conversion electrons per gamma-ray photon.
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(d) Higher $Z$ increases the electron wavefunction density at the nuclear volume, enhancing the coupling.
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(b) The conversion electron receives the transition energy minus the binding energy of the ejected shell.
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(b) The $K/L$ ratio depends primarily on multipole character, making it a diagnostic tool for assigning multipolarities.
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(b) A nuclear isomer is a metastable excited state, typically with $t_{1/2} > 100$ ns.
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(b) The $1/2^- \to 9/2^+$ transition requires M4 multipolarity ($\Delta J = 4$, parity change), which has a very slow transition rate.
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(c) The 140.5 keV energy and 6-hour half-life are optimally matched to gamma camera detection and clinical workflow.
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(b) The Mossbauer effect is recoilless resonance emission and absorption in solids.
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(b) $E_R = E_\gamma^2/(2Mc^2) = (14413)^2/(2 \times 57 \times 931.5 \times 10^6) \approx 1.96 \times 10^{-3}$ eV.
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(c) $f = \exp(-k^2\langle x^2\rangle)$: smaller $E_\gamma$ (smaller $k$) and higher $\Theta_D$ (smaller $\langle x^2 \rangle$) both increase $f$.
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(b) $E_R \propto E_\gamma^2$, so the 1332 keV gamma ray has $E_R \sim 12$ eV, far exceeding $k_B\Theta_D$, giving $f \approx 0$.
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(b) The gravitational redshift is $\Delta E/E = gh/c^2$ from general relativity.
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(b) The isomer shift depends on the s-electron contact density, which varies with oxidation state and bonding.
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(b) Magnetic splitting produces $2I+1$ sublevels for each nuclear state; the $\Delta m = 0, \pm 1$ selection rule allows exactly 6 transitions between the $I = 1/2$ and $I = 3/2$ states.
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(b) Recoilless emission/absorption means the linewidth is limited only by the natural linewidth $\Gamma = \hbar/\tau$, which is extraordinarily narrow for nuclear excited states.
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(d) E0 transitions are specific to transitions between same-$J$, same-$\pi$ states and are forbidden for photon emission. They are not "the dominant mechanism for all nuclear transitions."