Chapter 12 Quiz — Radioactivity Fundamentals
Instructions: Select the best answer for each question. Each question has exactly one correct answer.
Q1. The fundamental assumption of the radioactive decay law is that each nucleus:
(a) Decays after a predetermined time set at its creation (b) Has a constant probability $\lambda \, dt$ of decaying in time $dt$, independent of its age (c) Decays when it accumulates enough thermal energy to overcome the potential barrier (d) Has a probability of decaying that increases linearly with time
Q2. A sample initially contains $N_0$ radioactive atoms. After exactly 4 half-lives, how many atoms remain undecayed?
(a) $N_0 / 4$ (b) $N_0 / 8$ (c) $N_0 / 16$ (d) $N_0 / 32$
Q3. The mean life $\tau$ of a radioactive isotope is related to its half-life $t_{1/2}$ by:
(a) $\tau = t_{1/2}$ (b) $\tau = t_{1/2} / 2$ (c) $\tau = t_{1/2} / \ln 2 \approx 1.443 \, t_{1/2}$ (d) $\tau = t_{1/2} \times \ln 2 \approx 0.693 \, t_{1/2}$
Q4. One curie (Ci) is defined as:
(a) The activity of 1 gram of $^{238}$U (b) $3.7 \times 10^{10}$ decays per second (c) 1 decay per second (d) The activity that delivers 1 rad of dose per hour
Q5. Which isotope has the highest specific activity (activity per gram)?
(a) $^{238}$U ($t_{1/2} = 4.468 \times 10^9$ yr) (b) $^{226}$Ra ($t_{1/2} = 1600$ yr) (c) $^{60}$Co ($t_{1/2} = 5.271$ yr) (d) $^{3}$H ($t_{1/2} = 12.32$ yr)
Q6. In a two-member decay chain (parent $\to$ daughter $\to$ stable), secular equilibrium occurs when:
(a) The parent half-life is much shorter than the daughter half-life (b) The parent and daughter half-lives are equal (c) The parent half-life is much longer than the daughter half-life (d) Both parent and daughter are stable
Q7. At secular equilibrium in a decay chain, which statement is true?
(a) All members have the same number of atoms $N$ (b) All members have the same activity $A$ (c) All members have the same decay constant $\lambda$ (d) All members have the same half-life
Q8. In the transient equilibrium regime, the daughter activity is:
(a) Less than the parent activity (b) Equal to the parent activity (c) Greater than the parent activity, and both decay at the parent's rate (d) Greater than the parent activity, and both decay at the daughter's rate
Q9. If $^{212}$Bi can decay by $\alpha$ (36%) or $\beta^-$ (64%), and its total half-life is 60.55 min, what is the partial half-life for $\alpha$ decay?
(a) 21.8 min (b) 38.8 min (c) 60.55 min (d) 168 min
Q10. In carbon-14 dating, the "clock starts" when:
(a) The $^{14}$C atom is produced in the atmosphere (b) The organism dies and stops exchanging carbon with the environment (c) The sample is placed in the detector (d) The $^{14}$C/$^{12}$C ratio reaches $1.2 \times 10^{-12}$
Q11. The practical age range of $^{14}$C dating is approximately:
(a) 10 to 1,000 years (b) 300 to 50,000 years (c) 100,000 to 1 million years (d) 1 million to 4.5 billion years
Q12. In U-Pb dating, the concordia diagram plots:
(a) $^{206}$Pb/$^{238}$U vs. $^{207}$Pb/$^{235}$U, with concordant ages falling on a curve (b) $^{238}$U/$^{206}$Pb vs. time (c) Total lead content vs. total uranium content (d) $^{206}$Pb/$^{204}$Pb vs. $^{207}$Pb/$^{204}$Pb
Q13. The four natural radioactive series are classified by $A \bmod 4$ because:
(a) Proton number changes by 4 in each decay (b) $\alpha$ decay changes $A$ by 4, and $\beta$ decay does not change $A$ (c) There are four types of radioactive decay (d) Neutron number is always divisible by 4 in heavy nuclei
Q14. The neptunium series ($4n+1$) is "extinct" because:
(a) $^{237}$Np is too stable to decay (b) The series never existed in nature (c) The parent $^{237}$Np ($t_{1/2} = 2.14 \times 10^6$ yr) has decayed away over the 4.6 Gyr age of the Solar System (d) Neptunium-237 was only created in nuclear reactors
Q15. $^{99\text{m}}$Tc ($t_{1/2} = 6.006$ h) is the most widely used radioisotope in nuclear medicine. A hospital receives a 20 mCi source at 8:00 AM. The activity at 2:00 PM the same day is approximately:
(a) 20 mCi (b) 15 mCi (c) 10 mCi (d) 5 mCi
Q16. A sample of pure $^{226}$Ra is sealed in a container. After 30 days, the $^{222}$Rn activity will be approximately:
(a) Zero — radon has not had time to form (b) About 50% of the radium activity (c) Approximately equal to the radium activity (d) Much greater than the radium activity
Q17. The Rb-Sr isochron method eliminates the problem of unknown initial daughter content by:
(a) Assuming all initial $^{87}$Sr has decayed (b) Measuring multiple cogenetic samples with different Rb/Sr ratios and fitting a line (c) Using a standard with known $^{87}$Sr content (d) Only analyzing samples with zero initial strontium
Q18. The Bateman equations describe:
(a) The probability of nuclear fission (b) The populations of members in a radioactive decay chain as functions of time (c) The binding energy as a function of mass number (d) The angular distribution of emitted radiation
Answer Key
| Q | Answer | Brief Explanation |
|---|---|---|
| 1 | (b) | The constant probability postulate is the foundation of the decay law. |
| 2 | (c) | $N = N_0 (1/2)^4 = N_0/16$. |
| 3 | (c) | $\tau = t_{1/2}/\ln 2$; the mean life is always $\approx 44\%$ longer than the half-life. |
| 4 | (b) | 1 Ci = $3.7 \times 10^{10}$ Bq, originally based on 1 g of $^{226}$Ra. |
| 5 | (d) | Tritium has the lightest mass and a relatively short half-life, maximizing $\lambda N_A / M$. |
| 6 | (c) | Secular equilibrium requires $t_{1/2,\text{parent}} \gg t_{1/2,\text{daughter}}$. |
| 7 | (b) | At secular equilibrium, all activities are equal: $A_1 = A_2 = \cdots$. |
| 8 | (c) | In transient equilibrium, $A_{\text{daughter}} = A_{\text{parent}} \times \lambda_2 / (\lambda_2 - \lambda_1) > A_{\text{parent}}$. |
| 9 | (d) | $t_{1/2,\alpha} = t_{1/2}/BR_\alpha = 60.55/0.36 = 168$ min. |
| 10 | (b) | Death stops carbon exchange, locking in the $^{14}$C/$^{12}$C ratio. |
| 11 | (b) | $^{14}$C dating works from a few hundred to ~50,000 years. |
| 12 | (a) | The concordia plots the two U-Pb ratios; concordant ages lie on the curve. |
| 13 | (b) | $\alpha$ decay removes 4 from $A$; $\beta$ decay preserves $A$. |
| 14 | (c) | $4.6 \times 10^9 / 2.14 \times 10^6 \approx 2150$ half-lives — effectively zero remains. |
| 15 | (c) | 6 hours $= 1$ half-life, so $A = 20/2 = 10$ mCi. |
| 16 | (c) | 30 days $\approx 7.8$ Rn half-lives (3.82 d), so Rn is at $\sim$99.6% of secular equilibrium. |
| 17 | (b) | The isochron method uses multiple samples to determine both age and initial ratio. |
| 18 | (b) | Bateman (1910) derived the general solutions for sequential radioactive decay. |