Free MCAT-style practice, 4E
MCAT Nuclear Decay and Half-Life Practice Questions
Ten free MCAT-style practice questions on isotopes, alpha, beta and gamma decay, half-life calculations, mass defect, binding energy, fission and fusion, with full explanations.
Quick concept primer
A nucleus is described by its atomic number Z (protons) and mass number A (protons plus neutrons). Isotopes share Z but differ in neutron number, and an element's listed atomic mass is the abundance-weighted average of its isotopes. A nucleus has less mass than its separate nucleons; this mass defect, converted with E = mc2 (1 u corresponds to about 931.5 MeV), is the binding energy.
Unstable nuclei decay in ways that conserve both A and Z. Alpha decay emits a helium-4 nucleus (A falls by 4, Z by 2). Beta-minus decay turns a neutron into a proton (Z rises by 1). Positron emission and electron capture turn a proton into a neutron (Z falls by 1). Decay is first order: after each half-life, half of the remaining nuclei are left, so N = N0(1/2)t/thalf and the decay constant is k = ln 2 / thalf. Both fission of heavy nuclei and fusion of light nuclei release energy because the products sit closer to the iron-nickel peak of binding energy per nucleon.
Traps: beta-minus raises Z even though an electron leaves, gamma rays change no numbers, and compare stability by binding energy per nucleon, not total binding energy.
10 practice questions
Question 1
Radon-222 (Z = 86), a radioactive gas that can collect in basements, emits two alpha particles in succession. Which nucleus remains?
- Lead-214
- Polonium-218
- Bismuth-214
- Lead-218
Show answer and explanation
Answer: A. Lead-214
Each alpha particle carries away 4 nucleons and 2 protons. After two of them, A = 222 − 8 = 214 and Z = 86 − 4 = 82, which is lead, so the product is lead-214. Polonium-218 (Z = 84) is the product after only the first alpha decay, and lead-218 keeps the right element but forgets to lower the mass number twice.
Question 2
A patient receives a dose of a tracer with an initial activity of 800 MBq. The isotope has a half-life of about 6.0 hours, and biological clearance is ignored. What is the activity after 24 hours?
- 400 MBq
- 200 MBq
- 100 MBq
- 50 MBq
Show answer and explanation
Answer: D. 50 MBq
Twenty-four hours is 24/6.0 = 4 half-lives, so the activity falls by (1/2)4 = 1/16: 800/16 = 50 MBq. Halving four times gives 400, 200, 100 and then 50 MBq. The 200 MBq choice treats the loss as linear, a quarter per 6 hours, instead of exponential.
Question 3
A detector records the following count rates from a radioactive sample, corrected for background.
| Time (min) | Counts per minute |
|---|---|
| 0 | 4000 |
| 15 | 2828 |
| 30 | 2000 |
| 45 | 1414 |
What is the half-life of the sample?
- 15 min
- 30 min
- 45 min
- 60 min
Show answer and explanation
Answer: B. 30 min
The count rate falls from 4000 to 2000, exactly half, between 0 and 30 minutes, and from 2828 to 1414 between 15 and 45 minutes, so the half-life is 30 minutes. In 15 minutes the rate falls by a factor of 1/√2, about 0.71, not 1/2, so 15 minutes is half a half-life.
Question 4
Fluorine-18 (Z = 9), used in PET imaging, decays by positron emission. What is the daughter nucleus?
- Neon-18
- Oxygen-18
- Fluorine-17
- Nitrogen-14
Show answer and explanation
Answer: B. Oxygen-18
Positron emission converts a proton into a neutron, so Z falls by 1 while A stays the same: Z = 8, A = 18, which is oxygen-18. Neon-18 would come from raising Z, the beta-minus direction. The emitted positron later annihilates with an electron, producing the paired gamma photons that a PET scanner detects.
Question 5
A free proton has a mass of 1.0073 u, a free neutron 1.0087 u, and a deuteron (one proton plus one neutron) 2.0136 u. Using 1 u = 931.5 MeV, what is the binding energy of the deuteron?
- 0.0024 MeV
- 1.1 MeV
- 2.2 MeV
- 1876 MeV
Show answer and explanation
Answer: C. 2.2 MeV
The mass defect is (1.0073 + 1.0087) − 2.0136 = 0.0024 u. Multiplying by 931.5 MeV per u gives a binding energy of about 2.2 MeV. The 1.1 MeV choice is the binding energy per nucleon, 0.0024 MeV leaves the defect in mass units, and 1876 MeV is roughly the deuteron's total mass energy.
Question 6
Each of the following statements about radioactive decay is correct EXCEPT:
- Alpha particles are stopped by a sheet of paper or the outer layer of skin
- Beta-minus decay converts a neutron into a proton inside the nucleus
- Electron capture lowers the atomic number by one
- Gamma emission lowers the atomic number by one
Show answer and explanation
Answer: D. Gamma emission lowers the atomic number by one
A gamma ray is a high-energy photon with no mass number and no charge, so gamma emission changes neither A nor Z; it only lets an excited nucleus drop to a lower energy state. Alpha particles are heavy and strongly ionizing, so they penetrate very little. Beta-minus decay turns a neutron into a proton, and electron capture turns a proton into a neutron, lowering Z.
Question 7
Why can both the splitting of a uranium nucleus and the joining of two hydrogen isotopes into helium release energy?
- In both cases the products have a higher binding energy per nucleon than the reactants
- In both cases the total number of nucleons decreases
- Fission destroys protons and fusion creates neutrons
- Both processes lower the strong nuclear force between nucleons
Show answer and explanation
Answer: A. In both cases the products have a higher binding energy per nucleon than the reactants
Binding energy per nucleon peaks near iron and nickel. Splitting a heavy nucleus and fusing light nuclei both move the products toward that peak, so the products are more tightly bound and have less total mass; the mass difference is released as energy. Nucleons are conserved in both reactions, so B is false, and protons are not destroyed in fission.
Question 8
An element has two stable isotopes with masses of 10.0 u and 11.0 u. Its average atomic mass is 10.8 u. What percentage of its atoms are the 11.0 u isotope?
- 20 percent
- 50 percent
- 80 percent
- 92 percent
Show answer and explanation
Answer: C. 80 percent
Let x be the fraction of the heavier isotope: 10.0(1 − x) + 11.0x = 10.8, so 10.0 + x = 10.8 and x = 0.80, or 80 percent. A quick check: the average lies much closer to 11.0 than to 10.0, so the heavier isotope must be the majority. The 20 percent choice is the abundance of the lighter isotope.
Question 9
The activity of a radioactive sample falls to 12.5 percent of its starting value in 36 days. What is its decay constant?
- 0.019 per day
- 0.058 per day
- 0.083 per day
- 0.17 per day
Show answer and explanation
Answer: B. 0.058 per day
12.5 percent is (1/2)3, so 36 days is three half-lives and the half-life is 12 days. The first-order decay constant is k = ln 2 / thalf = 0.693/12, about 0.058 per day. Using 36 days as the half-life gives 0.019, and 1/12 = 0.083 forgets the ln 2 factor.
Question 10
Oxygen-16 contains 8 protons and 8 neutrons, while mercury-202 contains 80 protons and 122 neutrons. Why do stable nuclei of high atomic number hold a surplus of neutrons?
- Neutrons bind to neighbors through the short-range nuclear force but feel no Coulomb push from the protons
- A neutron has a slight negative charge that partly neutralizes the protons
- Nuclear attraction exists only in proton-neutron pairs, so each proton needs its own neutron partner
- Additional neutrons offset the combined mass of the orbiting electrons
Show answer and explanation
Answer: A. Neutrons bind to neighbors through the short-range nuclear force but feel no Coulomb push from the protons
Proton-proton repulsion acts across the whole nucleus, while the strong force acts only between near neighbors. As Z grows, repulsion grows faster than the attraction among protons, so heavy nuclei need surplus neutrons, which feel the strong force but carry no charge. Neutrons are uncharged (B is false), and the nuclear force also acts between neutron pairs and proton pairs (C is false).