MCAT Chemical and Physical Foundations of Biological Systems Quiz: 4e Nuclear Decay Radioactivity
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4e Nuclear Decay RadioactivityQuestion 1 of 20

A radiopharmacy prepares 125I^{125}\text{I} seeds for brachytherapy; 125I^{125}\text{I} decays by electron capture to 125Te^{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?

Seed 2 has half the activity of Seed 1 because one additional half-life has elapsed.
Seed 2 has the same activity as Seed 1 because electron capture does not change mass number.
Seed 2 has twice the activity of Seed 1 because the decay constant is unchanged.
Seed 2 has one-quarter the activity of Seed 1 because one additional half-life has elapsed.
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MCAT Chemical and Physical Foundations of Biological Systems Quiz

MCAT Chemical and Physical Foundations of Biological Systems Quiz: 4e Nuclear Decay Radioactivity

Practice 4e Nuclear Decay Radioactivity in MCAT Chemical and Physical Foundations of Biological Systems with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on 4e Nuclear Decay Radioactivity, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A radiopharmacy prepares 125I^{125}\text{I} seeds for brachytherapy; 125I^{125}\text{I} decays by electron capture to 125Te^{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?

  1. Seed 2 has half the activity of Seed 1 because one additional half-life has elapsed. (correct answer)
  2. Seed 2 has the same activity as Seed 1 because electron capture does not change mass number.
  3. Seed 2 has twice the activity of Seed 1 because the decay constant is unchanged.
  4. Seed 2 has one-quarter the activity of Seed 1 because one additional half-life has elapsed.

Explanation: 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.

Question 2

A researcher labels red blood cells with 51Cr^{51}\text{Cr} to track cell survival. 51Cr^{51}\text{Cr} decays primarily by electron capture to 51V^{51}\text{V} with a half-life of 27.7 days (decay constant λ≈2.9×10−7 s−1\lambda \approx 2.9\times 10^{-7}\ \text{s}^{-1}). Which statement best describes the decay process illustrated?

  1. A proton converts to a neutron, decreasing atomic number by 1 while mass number remains unchanged. (correct answer)
  2. A neutron converts to a proton, increasing atomic number by 1 while mass number remains unchanged.
  3. The nucleus emits a helium nucleus, decreasing mass number by 4 and atomic number by 2.
  4. The nucleus emits a gamma photon, decreasing mass number by 1 while atomic number is unchanged.

Explanation: 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.

Question 3

A laboratory prepares a sealed standard containing 60Co^{60}\text{Co}, which decays by β−\beta^- emission to an excited state of 60Ni^{60}\text{Ni} followed by gamma emission. The half-life of 60Co^{60}\text{Co} is 5.27 y. Based on the decay model, what outcome is most likely regarding the sequence of emissions?

  1. Beta emission changes the element, and subsequent gamma emission relaxes the daughter nucleus without changing its identity (correct answer)
  2. Gamma emission changes the element, and subsequent beta emission relaxes the nucleus without changing identity
  3. Beta emission decreases mass number by 4, and gamma emission decreases mass number by 1
  4. Gamma emission must occur before beta emission because photons are faster than electrons

Explanation: This question tests understanding of sequential nuclear decay processes. In ⁶⁰Co decay, β- emission occurs first, converting ⁶⁰Co to ⁶⁰Ni* (excited state) by changing a neutron to proton, thus changing the element. The excited ⁶⁰Ni* then undergoes gamma decay to ground state ⁶⁰Ni, releasing energy without changing nuclear composition. Choice A correctly describes beta emission changing the element followed by gamma emission relaxing the daughter nucleus. Choice B reverses the sequence impossibly, choice C gives incorrect mass changes, and choice D incorrectly relates emission order to particle speed. When analyzing decay chains, beta decay changes element identity while subsequent gamma decay only changes energy state.

Question 4

A research lab labels antibodies with 131I^{131}\text{I} for a targeted therapy model. Assume 131I^{131}\text{I} undergoes β−\beta^- decay to 131Xe^{131}\text{Xe}. The decay constant is λ=1.0×10−6 s−1\lambda = 1.0\times 10^{-6}\ \text{s}^{-1}. Based on the decay model, what is most consistent with conservation laws for the nuclear reaction?

  1. Mass number remains 131 and atomic number increases by 1, with an electron emitted (correct answer)
  2. Mass number decreases by 1 and atomic number remains constant, with a neutron emitted
  3. Mass number decreases by 4 and atomic number decreases by 2, with an alpha particle emitted
  4. Mass number increases by 1 and atomic number decreases by 1, with a positron emitted

Explanation: This question tests understanding of beta-minus decay and conservation laws. In β- decay of ¹³¹I to ¹³¹Xe, a neutron converts to a proton: n → p + e- + ν̄e. This increases atomic number from 53 (iodine) to 54 (xenon) while mass number remains 131, conserving baryon number. An electron is emitted to conserve charge. Choice A correctly describes mass number conservation at 131, atomic number increase by 1, and electron emission. Choice B incorrectly suggests neutron emission, choice C describes alpha decay, and choice D is physically impossible. When verifying nuclear reactions, check that mass number, charge, and baryon number are conserved on both sides of the equation.

Question 5

A patient receives a therapeutic radionuclide that decays by alpha emission. The clinician notes that alpha particles have high linear energy transfer and short range in tissue. Which statement best describes what must be true about the nuclear change in alpha decay?

  1. The nucleus loses 2 protons and 2 neutrons, decreasing atomic number by 2 and mass number by 4 (correct answer)
  2. The nucleus gains 2 protons and 2 neutrons, increasing atomic number by 2 and mass number by 4
  3. The nucleus loses 1 proton and gains 1 neutron, decreasing atomic number by 1 with no mass change
  4. The nucleus remains unchanged because alpha particles are emitted from the electron cloud

Explanation: This question tests understanding of alpha decay and its nuclear changes. Alpha particles are helium-4 nuclei (²He⁴) containing 2 protons and 2 neutrons. When emitted, the parent nucleus loses these 4 nucleons, decreasing atomic number by 2 and mass number by 4. This explains alpha particles' high mass and charge, leading to high linear energy transfer and short tissue range. Choice A correctly describes the loss of 2 protons and 2 neutrons with corresponding decreases in atomic and mass numbers. Choice B incorrectly suggests gaining nucleons, choice C describes a different process, and choice D incorrectly places alpha emission outside the nucleus. When analyzing alpha decay, remember the emitted particle is a complete helium nucleus.

Question 6

A PET tracer sample contains 18F^{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)?

  1. About 25% remains because two half-lives have elapsed (correct answer)
  2. About 50% remains because half-life depends on initial activity
  3. About 75% remains because two half-lives remove only one quarter
  4. About 12.5% remains because three half-lives have elapsed

Explanation: 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.

Question 7

A sealed source undergoes gamma decay from an excited nuclear state with decay constant λ=5.0×10−2 s−1\lambda = 5.0\times 10^{-2}\ \text{s}^{-1}. Based on the decay model, what outcome is most likely regarding the emitted radiation and nuclear composition?

  1. A photon is emitted and the nucleus retains the same numbers of protons and neutrons (correct answer)
  2. An electron is emitted and the nucleus gains one proton while losing one neutron
  3. A helium nucleus is emitted and the nucleus loses two protons and two neutrons
  4. A positron is emitted and the nucleus gains one neutron while losing one proton

Explanation: This question tests understanding of gamma decay characteristics. Gamma decay involves emission of a high-energy photon from an excited nucleus transitioning to a lower energy state. Unlike particle emission, gamma decay doesn't change the number of protons or neutrons, so both atomic number and mass number remain constant. The nucleus retains its identity but loses energy. Choice A correctly describes photon emission with unchanged nuclear composition. Choice B describes β- decay, choice C describes alpha decay, and choice D describes β+ decay. When analyzing gamma decay, remember it's purely an energy transition without changing nuclear constituents.

Question 8

A radiotracer undergoes β−\beta^- decay in vivo. The emitted electron is detected indirectly via downstream instrumentation. Which statement best describes the decay process illustrated in terms of nucleon number conservation?

  1. Mass number is conserved while atomic number increases by 1 due to neutron-to-proton conversion (correct answer)
  2. Mass number decreases by 1 because the emitted electron carries away one nucleon
  3. Atomic number decreases by 1 because a proton converts to a neutron and emits an electron
  4. Both mass number and atomic number remain unchanged because beta particles are photons

Explanation: This question tests understanding of beta-minus decay and nucleon conservation. In β- decay, a neutron converts to a proton (n → p + e- + ν̄e), increasing atomic number by 1 while mass number remains constant since the total number of nucleons (protons + neutrons) is unchanged. The emitted electron is not a nucleon and doesn't affect mass number. Choice A correctly describes mass number conservation with atomic number increase due to neutron-to-proton conversion. Choice B incorrectly suggests mass number decrease, choice C reverses the process, and choice D incorrectly identifies beta particles as photons. When analyzing β- decay, remember that nucleon number (mass number) is conserved while proton number (atomic number) increases by 1.

Question 9

In a nuclear medicine imaging study, a patient receives an injection of 18F^{18}\text{F}-FDG. The radionuclide 18F^{18}\text{F} decays by positron emission to 18O^{18}\text{O}. The half-life of 18F^{18}\text{F} is 110 min. Based on the decay model, what outcome is most consistent with this process at the nuclear level?

  1. The atomic number decreases by 1 while the mass number remains unchanged, consistent with conversion of a proton to a neutron. (correct answer)
  2. The atomic number increases by 1 while the mass number remains unchanged, consistent with conversion of a neutron to a proton.
  3. The mass number decreases by 4 and the atomic number decreases by 2, consistent with emission of a helium nucleus.
  4. The nucleus emits a high-energy photon without changing atomic number or mass number, consistent with de-excitation only.

Explanation: 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.

Question 10

A hospital stores a vial of 131I^{131}\text{I} for thyroid ablation. 131I^{131}\text{I} decays by β−\beta^- emission with t1/2=8.0t_{1/2}=8.0 days. A physicist estimates the decay constant using λ=ln⁡2/t1/2\lambda = \ln 2 / t_{1/2}. Which value is most consistent with this model (units must match days)?

  1. λ≈0.087 day−1\lambda \approx 0.087\ \text{day}^{-1} (correct answer)
  2. λ≈0.69 day−1\lambda \approx 0.69\ \text{day}^{-1}
  3. λ≈0.125 day−2\lambda \approx 0.125\ \text{day}^{-2}
  4. λ≈11.5 day−1\lambda \approx 11.5\ \text{day}^{-1}

Explanation: This question tests understanding of nuclear decay and radioactivity, specifically calculating decay constants from half-life. Using the relationship λ = ln(2)/t₁/₂ with t₁/₂ = 8.0 days, we calculate λ = 0.693/8.0 ≈ 0.087 day⁻¹. Choice A correctly provides this value with appropriate units. Choice B gives approximately 0.69 day⁻¹, which would correspond to a half-life of about 1 day, not 8 days. When calculating decay constants, ensure the units of λ are reciprocal time units matching the half-life units, and remember that ln(2) ≈ 0.693 for quick estimates.

Question 11

A targeted alpha therapy (TAT) agent uses 223Ra^{223}\text{Ra}, which undergoes alpha decay as part of its decay chain. Considering only a single alpha decay event, which statement best describes the daughter nucleus relative to the parent 223Ra^{223}\text{Ra} nucleus?

  1. The daughter has mass number 219 and atomic number 86, reflecting loss of a 4He^{4}\text{He} nucleus. (correct answer)
  2. The daughter has mass number 223 and atomic number 87, reflecting conversion of a neutron to a proton.
  3. The daughter has mass number 222 and atomic number 87, reflecting emission of a positron.
  4. The daughter has mass number 223 and atomic number 88, reflecting emission of a gamma photon only.

Explanation: This question tests understanding of nuclear decay and radioactivity, specifically alpha decay characteristics. In alpha decay, the nucleus emits an alpha particle (⁴He nucleus) containing 2 protons and 2 neutrons. For ²²³Ra (radium, atomic number 88), alpha decay produces a daughter with mass number 219 (223 - 4) and atomic number 86 (88 - 2), which is radon (Rn). Choice A correctly describes these changes: mass number decreases by 4 and atomic number decreases by 2. Choice B incorrectly describes beta-minus decay, while choice D incorrectly suggests no change occurs. When analyzing alpha decay, always subtract 4 from the mass number and 2 from the atomic number to identify the daughter nucleus.

Question 12

In targeted radionuclide therapy, 131I^{131}\text{I} is used because it undergoes β−\beta^- decay to 131Xe^{131}\text{Xe}. A clinic models activity as A(t)=A0e−λtA(t)=A_0 e^{-\lambda t}. Provided: t1/2=8.0 dayst_{1/2}=8.0\ \text{days} and λ=8.66×10−2 day−1\lambda=8.66\times10^{-2}\ \text{day}^{-1}. Which statement best describes the decay process illustrated?

  1. A proton converts to a neutron, decreasing atomic number by 1 while mass number stays the same
  2. A neutron converts to a proton, increasing atomic number by 1 while mass number stays the same (correct answer)
  3. The nucleus emits a helium nucleus, decreasing mass number by 4 and atomic number by 2
  4. The nucleus emits a gamma photon, so both mass number and atomic number decrease

Explanation: 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.

Question 13

A researcher studies alpha-emitting 223Ra^{223}\text{Ra} used for bone metastasis therapy. The decay is modeled by N(t)=N0e−λtN(t)=N_0 e^{-\lambda t}. Data: t1/2=11.4 dayst_{1/2}=11.4\ \text{days} (so λ=6.08×10−2 day−1\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?

  1. Mass number decreases by 4 and atomic number decreases by 2 (correct answer)
  2. Mass number stays the same and atomic number increases by 1
  3. Mass number decreases by 1 and atomic number decreases by 1
  4. Mass number increases by 4 and atomic number increases by 2

Explanation: 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.

Question 14

A researcher uses 14C^{14}\text{C} dating on biological samples; 14C^{14}\text{C} decays by β−\beta^- emission. Data: t1/2=5730 yt_{1/2}=5730\ \text{y} (so λ=1.21×10−4 y−1\lambda=1.21\times10^{-4}\ \text{y}^{-1}). Based on β−\beta^- decay, what change occurs to the nucleus?

  1. Atomic number increases by 1 while mass number remains 14 (correct answer)
  2. Atomic number decreases by 1 while mass number remains 14
  3. Mass number decreases by 4 and atomic number decreases by 2
  4. Mass number decreases by 1 and atomic number remains unchanged

Explanation: 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 ¹⁴C, β⁻ decay increases atomic number by 1 while keeping mass number constant. Choice A is correct because it matches the neutron-to-proton conversion. 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.

Question 15

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 t1/2=5.0 dayst_{1/2}=5.0\ \text{days} (λ=1.39×10−1 day−1\lambda=1.39\times10^{-1}\ \text{day}^{-1}) and undergoes alpha decay. Which statement best describes the emitted particle and nuclear change?

  1. Emission of a 4He^4\text{He} nucleus; daughter has A−4A-4 and Z−2Z-2 (correct answer)
  2. Emission of an electron; daughter has AA and Z+1Z+1
  3. Emission of a positron; daughter has AA and Z+1Z+1
  4. Emission of a photon; daughter has A−1A-1 and ZZ

Explanation: 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.

Question 16

In an environmental decay scenario, a soil core contains 210Po^{210}\text{Po}, an alpha emitter. The half-life is 138 days (λ=5.02×10−3 day−1\lambda=5.02\times10^{-3}\ \text{day}^{-1}). Which statement best describes the directionality of the decay sequence for the nucleus?

  1. The daughter nucleus will have lower mass number and lower atomic number than the parent (correct answer)
  2. The daughter nucleus will have higher mass number and higher atomic number than the parent
  3. The daughter nucleus will have the same mass number but higher atomic number than the parent
  4. The daughter nucleus will have the same mass number and same atomic number as the parent

Explanation: 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 ²¹⁰Po alpha decay, the daughter has lower A and Z. Choice A is correct because alpha emission reduces both by 4 and 2, respectively. Choice C is incorrect as it suggests unchanged A but increased Z, like β⁻. When assessing decay scenarios, ensure the decay type aligns with given isotopic characteristics and half-life data. Cross-check decay products and process assumptions.

Question 17

A nuclear medicine application uses 99mTc^{99\text{m}}\text{Tc}, which emits a gamma photon when transitioning to 99Tc^{99}\text{Tc}. The half-life is 6.0 h (λ=3.21×10−2 h−1\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?

  1. It remains technetium because gamma emission does not change atomic number (correct answer)
  2. It becomes molybdenum because gamma emission decreases atomic number by 1
  3. It becomes ruthenium because gamma emission increases atomic number by 1
  4. It becomes a different element because gamma emission decreases mass number by 4

Explanation: 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.

Question 18

A lab technician computes half-life from a measured decay constant for a β+\beta^+ tracer: λ=6.30×10−3 min−1\lambda=6.30\times10^{-3}\ \text{min}^{-1}. Which value is most consistent with the model t1/2=0.693λt_{1/2}=\frac{0.693}{\lambda}?

  1. 11.0 min
  2. 55.0 min
  3. 110 min (correct answer)
  4. 220 min

Explanation: 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. Half-life is calculated as t½=0.693/λ for the β⁺ tracer. Choice C is correct because it computes 110 min accurately. Choice A is incorrect as it underestimates by a factor of 10. When assessing decay scenarios, ensure the decay type aligns with given isotopic characteristics and half-life data. Cross-check decay products and process assumptions.

Question 19

A clinician compares two therapeutic isotopes that both emit alpha particles. Isotope A has t1/2=11.4 dayst_{1/2}=11.4\ \text{days} and Isotope B has t1/2=3.82 dayst_{1/2}=3.82\ \text{days}. Both are administered at the same initial number of nuclei. Based on half-life alone, what outcome is most likely?

  1. Isotope B will have a larger decay constant and higher initial activity than Isotope A (correct answer)
  2. Isotope B will have a smaller decay constant and lower initial activity than Isotope A
  3. Both will have the same decay constant because both are alpha emitters
  4. Neither isotope's decay constant can be compared without knowing the daughter isotope

Explanation: 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. Shorter half-life means larger λ and faster initial decay for same number of nuclei. Choice A is correct because Isotope B has shorter t½, thus larger λ and higher initial activity. Choice B is incorrect as it reverses the relationship. When assessing decay scenarios, ensure the decay type aligns with given isotopic characteristics and half-life data. Cross-check decay products and process assumptions.

Question 20

A laboratory sample contains a radionuclide that decays by β−\beta^- emission. The parent is identified as 32P^{32}\text{P} (Z=15). Based on the decay principle, which daughter nuclide is most consistent with β−\beta^- decay (ignore any gamma emissions)?

  1. 32Si^{32}\text{Si} (Z=14)
  2. 32S^{32}\text{S} (Z=16) (correct answer)
  3. 28Al^{28}\text{Al} (Z=13)
  4. 36Cl^{36}\text{Cl} (Z=17)

Explanation: 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 β⁻ decay of ³²P (Z=15), Z increases to 16, producing ³²S. Choice B is correct because it identifies the daughter with same A and Z+1. Choice A is incorrect as it suggests Z-1. When assessing decay scenarios, ensure the decay type aligns with given isotopic characteristics and half-life data. Cross-check decay products and process assumptions.