Questions on radioactive decay and the decay constant

Question 1

Define the half-life of a radioactive isotope.

Accepted Answer

The average time taken for the number of undecayed nuclei in a sample to fall to half its initial value (1); equivalently, the time for the activity to halve (1).

Question 2

A sample contains 1.0 × 10²⁰ undecayed nuclei of an isotope with a decay constant of 0.02 s⁻¹. Calculate the activity of the sample.

Accepted Answer

A = λN = 0.02 × 1.0 × 10²⁰ (1) A = 2.0 × 10¹⁸ Bq (1)

Question 3

State the relationship between the half-life of an isotope and its decay constant.

Accepted Answer

T_½ = ln 2/λ (≈ 0.693/λ) (1).

Question 4

Radioactive decay is described as a random process. Explain what this means.

Accepted Answer

It is impossible to predict when a particular nucleus will decay, or which nucleus will decay next (1); each nucleus has a constant probability of decay per unit time, unaffected by temperature, pressure or chemical conditions (1).

Question 5

Define the activity of a radioactive source and state its SI unit.

Accepted Answer

Activity is the number of nuclei in the source that decay per unit time (1); its SI unit is the becquerel (Bq), one decay per second (1).

Question 6

A nucleus left in an excited state after a decay emits a gamma photon. State what happens to the proton number and the nucleon number of the nucleus, and explain why the photon is emitted.

Accepted Answer

Both the proton number and the nucleon number are unchanged (1). The photon carries away the energy released as the nucleus falls from an excited state (a higher nuclear energy level) to a lower one (1).

Question 7

Technetium-99m is injected into a patient as a medical tracer. It emits gamma radiation only, from an excited nuclear state, and has a half-life of 6 hours. Explain why each of these two properties makes it suitable for this use.

Accepted Answer

Gamma radiation is penetrating, so the photons escape from the body to reach the camera, and it is weakly ionising, so little energy is deposited in the patient's tissue (1). A half-life of 6 hours is long enough for the scan to be carried out but short enough for the activity to fall to a negligible level within a day or two, limiting the dose the patient receives (1).

Question 8

An isotope has a half-life of 6.0 hours. Calculate its decay constant in hour⁻¹.

Accepted Answer

λ = ln 2/T_½ = 0.693/6.0 (1) λ = 0.116 hour⁻¹ (1)

Question 9

A sample initially contains 8.0 × 10²⁰ undecayed nuclei of an isotope with a half-life of 5.0 days. Calculate the number of undecayed nuclei remaining after 15 days.

Accepted Answer

15 days = 3 half-lives (1) N = N₀/2³ = (8.0 × 10²⁰)/8 (1) N = 1.0 × 10²⁰ nuclei (1)

Question 10

A source has an initial activity of 400 Bq. Calculate its activity after two half-lives.

Accepted Answer

A = A₀/2² = 400/4 (1) A = 100 Bq (1)

Question 11

A sample contains 5.0 × 10¹⁸ undecayed nuclei of an isotope with a half-life of 8.0 s. Calculate the activity of the sample.

Accepted Answer

λ = ln 2/8.0 = 0.0866 s⁻¹ (1) A = λN = 0.0866 × 5.0 × 10¹⁸ (1) A = 4.33 × 10¹⁷ Bq (1)

Question 12

A sealed medical source contains 4.2 μg of iodine-131, which has a half-life of 8.0 days and a molar mass of 131 g mol⁻¹. Calculate the activity of the source.
N_A = 6.02 × 10²³ mol⁻¹, 1 day = 86 400 s

Accepted Answer

λ = ln 2/T_½ = 0.693/(8.0 × 86 400) = 1.00 × 10⁻⁶ s⁻¹ (1) N = (4.2 × 10⁻⁶/131) × 6.02 × 10²³ (1) N = 1.93 × 10¹⁶ nuclei (1) A = λN = 1.9 × 10¹⁰ Bq (1)

Question 13

A student measures the activity A of a sample at intervals and plots ln (A/Bq) against time. The graph is a straight line with gradient −0.046 minute⁻¹. State what the straight line shows about the decay, and determine the half-life of the sample.

Accepted Answer

A straight line of ln A against t shows the decay is exponential (1). The magnitude of the gradient equals the decay constant, so λ = 0.046 minute⁻¹ (1)
T_½ = ln 2/λ = 0.693/0.046 (1)
T_½ = 15 minutes (1)

Question 14

Cobalt-60, ⁶⁰₂₇Co, decays by beta-minus emission to an isotope of nickel, Ni. Write the equation for this decay.

Accepted Answer

⁶⁰₂₇Co → ⁶⁰₂₈Ni + ⁰₋₁β + antineutrino: the nickel isotope has nucleon number 60 and proton number 28 (1); the beta particle is accompanied by an (electron) antineutrino (1).

Question 15

A nuclide lies below the line of stability on a graph of neutron number against proton number, so it has too few neutrons for its number of protons. State two decay processes by which it could move towards the line, and describe the change inside the nucleus that both processes produce.

Accepted Answer

Beta-plus (positron) emission (1); electron capture (1). In both, a proton is converted into a neutron, so the proton number falls by one and the neutron number rises by one, moving the nuclide towards the line of stability (1).

Question 16

A student uses a Geiger tube to follow the decay of a source. With the source removed the tube records 25 counts per minute. With the source in place it records 985 counts per minute, and 45 minutes later it records 145 counts per minute. Determine the half-life of the source.

Accepted Answer

Subtract the background from each reading first, because the background does not decay (1)
Corrected rates: 985 − 25 = 960 and 145 − 25 = 120 counts per minute (1)
960/120 = 8 = 2³, so 3 half-lives have passed in 45 minutes (1)
T_½ = 45/3 = 15 minutes (1)

Question 17

A source has an initial activity of 600 Bq and a half-life of 10 hours. Calculate its activity after 25 hours.

Accepted Answer

λ = ln 2/10 = 0.0693 hour⁻¹ (1) A = A₀e^−λt = 600 × e^−0.0693 × 25 (1) A = 106 Bq (1)

Question 18

The activity of a source falls from 800 Bq to 50 Bq. The half-life of the isotope is 3.0 days. Determine the time taken.

Accepted Answer

A₀/A = 800/50 = 16 = 2⁴ (1) So 4 half-lives have elapsed (1) t = 4 × 3.0 = 12 days (1)

Question 19

A nuclide lies above the line of stability on a neutron-proton plot, so it has too many neutrons for its number of protons. Predict its likely mode of decay and explain your reasoning.

Accepted Answer

Beta-minus decay (1). A neutron changes into a proton, emitting an electron and an antineutrino (1); this reduces the neutron-to-proton ratio, moving the nuclide towards the line of stability (1).

Question 20

An engineer injects a radioactive tracer into an underground pipe to locate a leak. The tracer has a half-life of 6.0 hours and an activity of 6.4 × 10⁷ Bq when injected. The leak can be located only while the activity exceeds 1.0 × 10⁷ Bq. Determine the time available to find the leak.

Accepted Answer

λ = ln 2/6.0 = 0.116 hour⁻¹ (1) A = A₀e^−λt, so t = ln (A₀/A)/λ (1) t = ln (6.4 × 10⁷/(1.0 × 10⁷))/0.116 (1) t = 16 hours (1)

Question 21

Engineers add a radioactive tracer to the water in a buried pipe to find a leak, detecting the radiation from the surface 0.5 m above. Three tracers are available. P: alpha emitter, half-life 4.0 days; Q: gamma emitter, half-life 8.0 hours; R: gamma emitter, half-life 30 years. Deduce which tracer is most suitable, giving a reason for rejecting each of the others.

Accepted Answer

P is rejected: alpha particles are absorbed by a few centimetres of air, let alone half a metre of soil, so nothing would reach the surface detector (1). R is rejected: a 30 year half-life would leave the ground and the water supply contaminated for decades (1). Q is chosen: gamma radiation is penetrating enough to reach the surface through the soil (1); an 8.0 hour half-life lasts long enough to complete the survey but decays to a low level within a few days (1).

Question 22

Regulations require the activity of a container of radioactive waste to fall below one thousandth of its initial value before it may be moved for disposal. The activity is dominated by an isotope with a half-life of 30 years. Determine the minimum storage time.

Accepted Answer

After n half-lives the activity is A₀/2ⁿ (1) 2⁹ = 512 is not enough, but 2¹⁰ = 1024 > 1000, so 10 half-lives are needed (1) t = 10 × 30 = 300 years (1)

Question 23

Charcoal from an ancient hearth has a carbon-14 activity per gram of carbon that is 22% of the activity per gram of carbon in living wood. Carbon-14 has a half-life of 5730 years. Estimate the age of the charcoal.

Accepted Answer

While the tree was alive it exchanged carbon with the air, so its carbon-14 activity matched that of living wood; the activity has fallen only since the wood stopped exchanging carbon (1)
A/A₀ = 0.22, so (½)ⁿ = 0.22, where n is the number of half-lives (1)
n = ln (1/0.22)/ln 2 = 2.18 half-lives (1)
t = 2.18 × 5730 = 1.3 × 10⁴ years (1)

Question 24

A sample contains 8.0 × 10¹² undecayed nuclei of an isotope whose decay constant is 0.10 s⁻¹. A student models the decay in a spreadsheet from ΔN = −λNΔt, taking a time step of Δt = 1.0 s and starting each step from the number of undecayed nuclei left at the end of the step before.
Work through the first three steps of the model, giving the number of undecayed nuclei left after each one, and compare the value after 3.0 s with the number given by N = N₀e^−λt. Explain why the model differs from the equation in the way it does, and state one change to the model that would reduce the difference.

Accepted Answer

Step 1: ΔN = −λNΔt = −0.10 × 8.0 × 10¹² × 1.0 = −8.0 × 10¹¹, leaving N = 7.2 × 10¹² (1)
Step 2, from that new N: ΔN = −7.2 × 10¹¹, leaving N = 6.48 × 10¹²; step 3: ΔN = −6.48 × 10¹¹, leaving N = 5.83 × 10¹² after 3.0 s (1)
Equation: N = N₀e^−λt = 8.0 × 10¹² × e^{−0.10 × 3.0} = 5.93 × 10¹² (1)
The model is the lower of the two, by about 1.6% (1)
Across each step the model holds the decay rate at the value it had at the start of the step, while in the sample N, and with it the rate λN, falls throughout the step (1); each step therefore removes too many nuclei and the modelled count runs below the true exponential
Use a shorter time step, small enough that λΔt is small (0.10 s in place of 1.0 s), and repeat for correspondingly more steps (1)