Nuclear Decay and Radioactivity (4E) - Chem & Physics

1. In an in vitro assay, a peptide is labeled with $^{32}\text{P}$, which undergoes $\beta^-$ decay to $^{32}\text{S}$. The half-life is 14.3 days (decay constant $\lambda \approx 5.6\times 10^{-7}\ \text{s}^{-1}$). Based on the decay model, what outcome is most likely for the labeled molecule immediately after decay?
   Answer: The phosphorus atom becomes sulfur in the molecule, increasing atomic number by 1 while mass number stays 32. This question tests understanding of nuclear decay and radioactivity, specifically beta-minus decay and its chemical consequences. In beta-minus decay, a neutron converts to a proton, increasing the atomic number by 1. When P-32 (atomic number 15) undergoes beta-minus decay, it becomes S-32 (atomic number 16) while maintaining mass number 32. This nuclear transformation changes the chemical identity of the atom within the molecule from phosphorus to sulfur. Choice A is correct as it accurately describes this transmutation and the resulting atomic number increase. Choice D is incorrect because beta decay changes the nuclear composition and thus the element's identity, not just the electron configuration.
2. In a nuclear medicine imaging protocol, a patient receives an intravenous dose of $^{18}\text{F}$-FDG. The radionuclide undergoes $\beta^+$ decay: $^{18}\text{F} \rightarrow \,^{18}\text{O} + e^+ + \nu_e$. The decay constant is reported as $\lambda = 1.05\times 10^{-4}\ \text{s}^{-1}$. Based on this decay model, what outcome is most likely at the nuclear level?
   Answer: The atomic number decreases by 1 because a proton converts to a neutron, producing a positron This question tests understanding of nuclear decay and radioactivity, specifically beta-plus decay. In β+ decay, a proton in the nucleus converts to a neutron, emitting a positron (e+) and a neutrino (νe), which decreases the atomic number by 1 while the mass number remains constant. For ¹⁸F → ¹⁸O + e+ + νe, fluorine (Z=9) transforms to oxygen (Z=8), confirming the atomic number decreases by 1. Choice A correctly describes this process where a proton converts to a neutron with positron emission. Choice B incorrectly describes alpha decay, choice C describes β- decay (opposite process), and choice D describes gamma decay. When analyzing nuclear decay, verify that charge and mass numbers are conserved on both sides of the equation.
3. In targeted radionuclide therapy, $^{131}\text{I}$ is used because it undergoes $\beta^-$ decay to $^{131}\text{Xe}$. A clinic models activity as $A(t)=A_0 e^{-\lambda t}$. Provided: $t_{1/2}=8.0\ \text{days}$ and $\lambda=8.66\times10^{-2}\ \text{day}^{-1}$. Which statement best describes the decay process illustrated?
   Answer: A neutron converts to a proton, increasing atomic number by 1 while mass number stays the same This question tests understanding of nuclear decay and radioactivity (4E). Radioactive decay involves the transformation of an unstable nucleus into a more stable one, often emitting particles or radiation. Here, ¹³¹I decays by β⁻ emission, converting a neutron to a proton and emitting an electron. Choice B is correct because it accurately describes the increase in atomic number by 1 with unchanged mass number, matching β⁻ decay. Choice C is incorrect as it describes alpha decay, not β⁻ decay. When assessing decay scenarios, ensure the decay type aligns with given isotopic characteristics and half-life data. Cross-check decay products and process assumptions.
4. A research group compares two tracers for short imaging protocols. Tracer X has $t_{1/2}=2.0\ \text{h}$ and tracer Y has $t_{1/2}=8.0\ \text{h}$. Both decay by the same mode and follow first-order kinetics. Based on the decay model, what prediction can be made about their decay constants $\lambda$?
   Answer: $\lambda_X$ is four times $\lambda_Y$ because $\lambda$ is inversely proportional to half-life. This question tests understanding of nuclear decay and radioactivity, specifically the relationship between decay constant and half-life. The decay constant λ and half-life t₁/₂ are related by λ = ln(2)/t₁/₂, making them inversely proportional. For tracer X with t₁/₂ = 2.0 h, λₓ = ln(2)/2.0, while for tracer Y with t₁/₂ = 8.0 h, λᵧ = ln(2)/8.0. Since X has one-fourth the half-life of Y, it has four times the decay constant. Choice C correctly states that λₓ is four times λᵧ due to inverse proportionality. Choice A incorrectly suggests that shorter half-life means slower decay, when the opposite is true. Remember that larger decay constants correspond to faster decay and shorter half-lives.
5. In a nuclear medicine imaging study, a patient is injected with sodium iodide containing $^{131}\text{I}$, which undergoes $\beta^-$ decay to $^{131}\text{Xe}$. The isotope's half-life is 8.0 days (decay constant $\lambda \approx 1.0\times 10^{-6}\ \text{s}^{-1}$). Based on the decay model, what outcome is most likely for the daughter nucleus compared with the parent?
   Answer: The atomic number increases by 1 while the mass number remains unchanged. This question tests understanding of nuclear decay and radioactivity, specifically beta-minus decay. In beta-minus decay, a neutron in the nucleus converts to a proton while emitting an electron (beta particle) and an antineutrino. For the decay of I-131 to Xe-131, the mass number remains constant at 131 because the total number of nucleons is unchanged. However, the atomic number increases by 1 (from 53 for iodine to 54 for xenon) because a neutron has been converted to a proton. Choice B is correct as it accurately describes this increase in atomic number with unchanged mass number. Choice A is incorrect because it suggests the atomic number decreases, which would occur in beta-plus decay or electron capture, not beta-minus decay.
6. A hospital physicist compares two sealed calibration sources. Source X has $\lambda_X=0.010\ \text{day}^{-1}$ and Source Y has $\lambda_Y=0.040\ \text{day}^{-1}$. Based on the decay model, what prediction can be made about their half-lives?
   Answer: Source X has a longer half-life than Source Y because it has a smaller decay constant This question tests understanding of the inverse relationship between decay constant and half-life. Since t₁/₂ = ln(2)/λ, smaller decay constant means longer half-life. For Source X: t₁/₂ = 0.693/0.010 = 69.3 days; for Source Y: t₁/₂ = 0.693/0.040 = 17.3 days. Source X has the longer half-life due to its smaller decay constant. Choice A correctly identifies that smaller decay constant corresponds to longer half-life. Choice B reverses the relationship, choice C incorrectly claims identical half-lives, and choice D incorrectly introduces initial nuclei dependence. When comparing half-lives, remember that λ and t₁/₂ are inversely proportional.
7. A lab prepares a calibration standard of $^{99m}\text{Tc}$ for SPECT imaging. $^{99m}\text{Tc}$ undergoes gamma decay to $^{99}\text{Tc}$ with a half-life of 6.0 h. Based on the decay principle, which statement best describes what changes in the nucleus during this decay?
   Answer: Neither mass number nor atomic number changes; the nucleus transitions from an excited to a lower-energy state by photon emission. This question tests understanding of nuclear decay and radioactivity, specifically gamma decay characteristics. In gamma decay, an excited nucleus (indicated by 'm' for metastable in ⁹⁹ᵐTc) transitions to a lower energy state by emitting a gamma photon. Unlike alpha or beta decay, gamma emission involves no change in the number of protons or neutrons, so both mass number and atomic number remain constant. Choice C correctly states that neither mass number nor atomic number changes during gamma decay. Choice A describes positron emission, while choice D describes beta-minus decay, neither of which applies here. Remember that gamma decay is purely an energy transition without particle emission, distinguishing it from other decay modes that alter nuclear composition.
8. A detector is placed next to a sealed vial containing a pure gamma-emitting radionuclide with half-life 10 h. The detector counts rate is proportional to activity. Based on the decay model, what outcome is most likely for the count rate after 30 h, assuming constant geometry and efficiency?
   Answer: It is about $1/8$ of the initial rate because three half-lives have elapsed This question tests understanding of activity decay over multiple half-lives. With t₁/₂ = 10 h and elapsed time = 30 h, three half-lives have passed (30/10 = 3). Count rate, proportional to activity, follows: after 1 half-life → 1/2 initial; after 2 half-lives → 1/4 initial; after 3 half-lives → 1/8 initial. The general formula is rate(t) = rate(0) × $(1/2)^{t/t₁/₂}$. Choice A correctly identifies 1/8 of initial rate after three half-lives. Choice B incorrectly assumes linear decay, choice C undercounts half-lives, and choice D overcounts to four half-lives. When calculating count rates, remember they're directly proportional to activity and follow the same exponential decay.
9. In a nuclear medicine imaging study, a patient receives an injection of $^{18}\text{F}$-FDG. The radionuclide $^{18}\text{F}$ decays by positron emission to $^{18}\text{O}$. The half-life of $^{18}\text{F}$ is 110 min. Based on the decay model, what outcome is most consistent with this process at the nuclear level?
   Answer: The atomic number decreases by 1 while the mass number remains unchanged, consistent with conversion of a proton to a neutron. This question tests understanding of nuclear decay and radioactivity, specifically positron emission. In positron emission, a proton in the nucleus converts to a neutron while emitting a positron (β+) and a neutrino. For ¹⁸F decaying to ¹⁸O, the mass number remains 18 (unchanged) while the atomic number decreases from 9 (fluorine) to 8 (oxygen). Choice A correctly describes this process: atomic number decreases by 1 while mass number remains unchanged, consistent with proton-to-neutron conversion. Choice B incorrectly describes beta-minus decay where a neutron converts to a proton. When analyzing positron emission, remember that the atomic number decreases while mass number stays constant, and verify the daughter nucleus has one fewer proton than the parent.
10. A nuclear medicine application uses $^{99\text{m}}\text{Tc}$, which emits a gamma photon when transitioning to $^{99}\text{Tc}$. The half-life is 6.0 h ($\lambda=3.21\times10^{-2}\ \text{h}^{-1}$). Based on the decay process, what outcome is most likely concerning the chemical identity of the atom after decay?
    Answer: It remains technetium because gamma emission does not change atomic number This question tests understanding of nuclear decay and radioactivity (4E). Radioactive decay involves the transformation of an unstable nucleus into a more stable one, often emitting particles or radiation. For ⁹⁹ᵐTc, gamma emission does not change the element. Choice A is correct because Z remains unchanged, keeping it technetium. Choice B is incorrect as it falsely claims a decrease in Z. When assessing decay scenarios, ensure the decay type aligns with given isotopic characteristics and half-life data. Cross-check decay products and process assumptions.
11. A researcher labels red blood cells with $^{51}\text{Cr}$ to track cell survival. $^{51}\text{Cr}$ decays primarily by electron capture to $^{51}\text{V}$ with a half-life of 27.7 days (decay constant $\lambda \approx 2.9\times 10^{-7}\ \text{s}^{-1}$). Which statement best describes the decay process illustrated?
    Answer: A proton converts to a neutron, decreasing atomic number by 1 while mass number remains unchanged. This question tests understanding of nuclear decay and radioactivity, specifically electron capture. In electron capture, a proton in the nucleus captures an inner orbital electron and converts to a neutron, emitting a neutrino. When Cr-51 undergoes electron capture to V-51, the atomic number decreases by 1 (from 24 for chromium to 23 for vanadium) while the mass number remains at 51. Choice A is correct as it accurately describes the conversion of a proton to a neutron and the resulting decrease in atomic number. Choice B is incorrect because it describes beta-minus decay (neutron to proton conversion), which is the opposite process.
12. A radiopharmacy prepares $^{125}\text{I}$ seeds for brachytherapy; $^{125}\text{I}$ decays by electron capture to $^{125}\text{Te}$. The half-life is 59 days. Two identical sealed seeds are stored: Seed 1 is stored for 59 days; Seed 2 is stored for 118 days. What prediction can be made about the decay rate (activity) of Seed 2 relative to Seed 1 at the time of use, assuming identical initial activity?
    Answer: Seed 2 has half the activity of Seed 1 because one additional half-life has elapsed. This question tests understanding of nuclear decay and radioactivity, specifically the concept of half-life and activity calculations. After one half-life (59 days), Seed 1 will have half its original activity. After two half-lives (118 days), Seed 2 will have undergone two halvings: (1/2) × (1/2) = 1/4 of its original activity. Since both seeds started with identical initial activity, Seed 2 will have half the activity of Seed 1 at their respective times of use. Choice A is correct because Seed 2 has experienced one additional half-life compared to Seed 1. Choice D is incorrect because it compares Seed 2's activity to the original activity rather than to Seed 1's activity at 59 days.
13. In a lab setup, a sealed vial contains a pure sample of a gamma-emitting isomer used as a timing standard. The measured decay constant is $\lambda=0.050\ \text{min}^{-1}$. Based on this value, what prediction can be made about its half-life?
    Answer: It is approximately $t_{1/2}=\frac{0.693}{0.050}\approx 14\ \text{min}$ This question tests understanding of nuclear decay and radioactivity (4E). Radioactive decay involves the transformation of an unstable nucleus into a more stable one, often emitting particles or radiation. Here, half-life is calculated from λ using t½=0.693/λ. Choice A is correct because it accurately computes ≈14 min. Choice B is incorrect as it inverts the formula. When assessing decay scenarios, ensure the decay type aligns with given isotopic characteristics and half-life data. Cross-check decay products and process assumptions.
14. A SPECT tracer decays by electron capture (EC), in which a proton in the nucleus converts to a neutron after capturing an inner-shell electron. The tracer has $t_{1/2}=13\ \text{h}$ ($\lambda=5.33\times10^{-2}\ \text{h}^{-1}$). Based on this decay process, what outcome is most likely for the daughter nucleus?
    Answer: Atomic number decreases by 1 while mass number stays constant This question tests understanding of nuclear decay and radioactivity (4E). Radioactive decay involves the transformation of an unstable nucleus into a more stable one, often emitting particles or radiation. In electron capture, a proton converts to a neutron, decreasing Z by 1 with A unchanged. Choice A is correct because it matches the nuclear change. Choice B is incorrect as it describes β⁻ decay. When assessing decay scenarios, ensure the decay type aligns with given isotopic characteristics and half-life data. Cross-check decay products and process assumptions.
15. A researcher studies alpha-emitting $^{223}\text{Ra}$ used for bone metastasis therapy. The decay is modeled by $N(t)=N_0 e^{-\lambda t}$. Data: $t_{1/2}=11.4\ \text{days}$ (so $\lambda=6.08\times10^{-2}\ \text{day}^{-1}$). Based on alpha decay, what outcome is most consistent for the daughter nucleus immediately after a decay event?
    Answer: Mass number decreases by 4 and atomic number decreases by 2 This question tests understanding of nuclear decay and radioactivity (4E). Radioactive decay involves the transformation of an unstable nucleus into a more stable one, often emitting particles or radiation. In this case, ²²³Ra undergoes alpha decay, emitting a helium nucleus. Choice A is correct because it describes the decrease in mass number by 4 and atomic number by 2, typical of alpha decay. Choice B is incorrect as it describes β⁻ decay, not alpha decay. When assessing decay scenarios, ensure the decay type aligns with given isotopic characteristics and half-life data. Cross-check decay products and process assumptions.
16. A PET tracer sample contains $^{18}\text{F}$ with half-life 110 min. The sample is transported for 220 min before use. Based on the decay model, what outcome is most likely for the remaining activity (ignoring biological clearance)?
    Answer: About 25% remains because two half-lives have elapsed This question tests understanding of radioactive decay over multiple half-lives. With t₁/₂ = 110 min and transport time = 220 min, exactly 2 half-lives have elapsed (220/110 = 2). After one half-life, 50% remains; after two half-lives, 25% remains. The formula is: fraction remaining = $(1/2)^{t/t₁/₂}$ = (1/2)² = 1/4 = 25%. Choice A correctly identifies that 25% remains after two half-lives. Choice B incorrectly ignores time dependence, choice C miscalculates the fraction, and choice D incorrectly counts three half-lives. When calculating remaining activity, always determine the number of half-lives as t/t₁/₂ and apply $(1/2)^n$.
17. A laboratory measures the decay of $^{32}\text{P}$, a $\beta^-$ emitter used in molecular biology labeling. The half-life is 14.3 days (so $\lambda=4.85\times10^{-2}\ \text{day}^{-1}$). Which statement best describes the nuclear change that accompanies $\beta^-$ decay?
    Answer: A neutron converts to a proton, so the atomic number increases by 1 while mass number stays constant This question tests understanding of nuclear decay and radioactivity (4E). Radioactive decay involves the transformation of an unstable nucleus into a more stable one, often emitting particles or radiation. For ³²P, β⁻ decay converts a neutron to a proton. Choice A is correct because it describes the increase in atomic number by 1 with constant mass number. Choice B is incorrect as it describes β⁺ decay instead. When assessing decay scenarios, ensure the decay type aligns with given isotopic characteristics and half-life data. Cross-check decay products and process assumptions.
18. In a PET facility, $^{11}\text{C}$-labeled tracer is synthesized and delivered to a scanner. $^{11}\text{C}$ decays by positron emission with half-life 20.3 min. A batch has activity 12 MBq at delivery. Based on the decay model, what activity is most likely 40.6 min after delivery (assume no biological clearance)?
    Answer: 3 MBq, because 40.6 min corresponds to two half-lives. This question tests understanding of nuclear decay and radioactivity, specifically activity calculations over multiple half-lives. With a half-life of 20.3 minutes, the elapsed time of 40.6 minutes represents exactly two half-lives (40.6 ÷ 20.3 = 2). After one half-life, activity decreases to 6 MBq (12 ÷ 2), and after two half-lives, it decreases to 3 MBq (6 ÷ 2). Choice A correctly identifies both the number of half-lives and the resulting activity. Choice D incorrectly suggests that positron emission increases activity, when in fact all radioactive decay decreases the number of parent nuclei over time. When calculating activity after multiple half-lives, divide the initial activity by 2ⁿ where n is the number of half-lives elapsed.
19. A clinician selects an alpha-emitting radionuclide for therapy and notes that alpha particles have low penetration in tissue but high ionization density. The isotope has $t_{1/2}=5.0\ \text{days}$ ($\lambda=1.39\times10^{-1}\ \text{day}^{-1}$) and undergoes alpha decay. Which statement best describes the emitted particle and nuclear change?
    Answer: Emission of a $^4\text{He}$ nucleus; daughter has $A-4$ and $Z-2$ This question tests understanding of nuclear decay and radioactivity (4E). Radioactive decay involves the transformation of an unstable nucleus into a more stable one, often emitting particles or radiation. For this alpha emitter, decay emits a ⁴He nucleus, reducing A by 4 and Z by 2. Choice A is correct because it describes the particle and nuclear change accurately. Choice B is incorrect as it describes β⁻ decay. When assessing decay scenarios, ensure the decay type aligns with given isotopic characteristics and half-life data. Cross-check decay products and process assumptions.
20. A quality-control lab measures the activity of a sealed $^{32}\text{P}$ source used for molecular biology assays. $^{32}\text{P}$ decays by $\beta^-$ emission with half-life 14.3 days. If the activity is 100 kBq today, which activity is most likely after 28.6 days (assume ideal exponential decay)?
    Answer: 25 kBq, because 28.6 days corresponds to two half-lives. This question tests understanding of nuclear decay and radioactivity, specifically activity calculations after multiple half-lives. With a half-life of 14.3 days, the elapsed time of 28.6 days represents exactly two half-lives (28.6 ÷ 14.3 = 2). Starting with 100 kBq, after one half-life the activity becomes 50 kBq, and after two half-lives it becomes 25 kBq. Choice A correctly identifies both the number of half-lives and the resulting activity. Choice C incorrectly calculates three half-lives, which would require 42.9 days. For exponential decay problems, first determine the number of half-lives by dividing elapsed time by the half-life, then apply the factor of (1/2)ⁿ to the initial activity.

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