3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
Model each nucleus as a sphere of radius R=R0A1/3 (the same R0 for all nuclei) with its charge Ze regarded as spread over the sphere. Define the surface charge density of a nucleus as its total charge divided by its surface area 4πR2. The ratio of the surface charge density of 92238U to that of 1327Al is closest to ______. Exactly one option is correct.
Show answer and solution
Answer: C
σ=4πR02A2/3Ze, so σAlσU=1392(23827)2/3=7.077×0.2343=1.66.
Traps: 0.80 is the ratio of volume charge densities (Z/A ratio, since R3∝A); 3.43 uses R instead of R2; 7.08 ignores the size difference.
Note: nuclear (mass) density is the same for all nuclei because R3∝A, but charge densities differ because Z/A is not constant.
Difficulty: HardQuestion 2Level 2 · ExamSingle correct
A detector records a fixed fraction of the decays occurring in a sample of a single radionuclide that decays to a stable daughter; background and dead time are negligible. Counting starts at t=0. The detector records 3000 counts between t=0 and t=1 h, and 2250 counts between t=1 h and t=3 h. The number of counts it records between t=3 h and t=4 h is ______. Exactly one option is correct.
Show answer and solution
Answer: B
Let x=e−λ(1 h) and let C be the counts that would be recorded over all time. Counts in [t1,t2] are C(e−λt1−e−λt2).
First hour: C(1−x)=3000. From 1 h to 3 h: C(x−x3)=Cx(1−x)(1+x)=2250.
Dividing: x(1+x)=0.75, whose positive root is x=0.5, so T1/2=1 h and C=6000.
Counts in [3,4] h: C(x3−x4)=6000(81−161)=375.
Trap: treating 2250/3000 as the ratio of successive one-hour counts gives a wrong half-life; the second window is two hours long.
Difficulty: HardQuestion 3Level 3 · AdvancedSingle correct
A 88226Ra nucleus at rest undergoes α decay to 86222Rn. Atomic masses: M(226Ra)=226.025410 u, M(222Rn)=222.017578 u, M(4He)=4.002603 u. Take 1 u=931.5 MeV/c2. In some decays the α particle is observed with kinetic energy 4.600 MeV; in these decays the daughter nucleus is left in an excited state, from which it then drops to its ground state by emitting a single γ photon. Treat the motion nonrelativistically, take the ratio of the masses of the α particle and the daughter nucleus equal to the ratio of their mass numbers, and neglect the recoil of the nucleus when the photon is emitted. The energy of the γ photon is closest to ______. Exactly one option is correct.
A188 keV
B185 keV
C271 keV
D95 keV
Show answer and solution
Answer: A
Ground-state Q value (atomic masses; electron masses cancel because the He atom carries two electrons): Q=(226.025410−222.017578−4.002603) u×931.5=0.005229×931.5=4.8708 MeV.
Momentum conservation from rest: pα=pRn, so the total kinetic energy is Kα(1+2224)=Kα222226.
For the excited-state branch, total kinetic energy =4.600×222226=4.6829 MeV, so the excitation energy is Eγ=4.8708−4.6829=0.1879 MeV≈188 keV.
Traps: 271 keV ignores the daughter's recoil (Q−4.600). 185 keV is the difference between the two α energies (Kα,0=Q⋅226222=4.7846 minus 4.600); differences in α energy must be multiplied by 226/222 because the recoil also shares the energy difference.
Difficulty: Peak challenge