Half-life and radioactive decay | OCR Gateway GCSE Physics Notes

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The key idea

Half-life is the time taken for the activity or number of undecayed nuclei to halve.

Half Life And Radioactive Decay

Use the labels to explain the scientific relationship shown.

Revision notes

The bit that matters

Short notes first. Learn the idea, then use the worked example and questions to check it properly.

Radioactive decay

Radioactive decay is a random process: it is impossible to predict which nucleus will decay next or exactly when.Activity is the rate at which nuclei in a source decay, measured in becquerels (Bq), where 1 Bq is one decay per second.Activity is measured with a Geiger-Muller tube and counter.As the unstable nuclei decay, the activity of a source falls over time.

Half-life

The half-life of a radioactive isotope is the average time taken for the number of undecayed nuclei in a sample to halve, or equivalently for the activity (count rate) to fall to half its starting value.Different isotopes have very different half-lives, from fractions of a second to billions of years.After each half-life the remaining amount halves again, falling to a half, then a quarter, then an eighth, and so on.

Calculating with half-life

To find how much remains, count the number of half-lives that have passed and halve the starting amount that many times.The fraction remaining after n half-lives is ( $\frac{1}{2}$ )ⁿ. For example, after 3 half-lives the remaining fraction is $\frac{1}{8}$ .You can also work backwards from the count rate to find how many half-lives have elapsed.

Nuclear decay and equations

When a nucleus emits an alpha particle its mass number falls by 4 and its atomic number falls by 2.When it emits a beta particle a neutron changes into a proton, so the mass number is unchanged but the atomic number increases by 1.Gamma emission carries away energy only and changes neither the mass number nor the atomic number.Nuclear equations must balance both numbers on each side.

Key terms

Definitions to learn

Activity

The rate of decay of a source, measured in becquerels (Bq).

Becquerel

One radioactive decay per second.

Half-life

The average time for half the undecayed nuclei to decay.

Count rate

The number of decays detected per second by a detector.

Random process

One where individual events cannot be predicted.

Worked example

A sample has an activity of 960 Bq. Its half-life is 3 days. Find the activity after 9 days.

Count the number of half-lives.

Halve the activity once for each half-life.

Final answer

120 Bq

Exam habit

Always divide total time by the half-life to find the number of halvings. Show each halving step.Never say 'after two half-lives, the sample is fully decayed' — it approaches zero but never reaches it.

Watch out

Radioactive decay is random for individual nuclei, even though a large sample follows a pattern.

Examiner tips

How to score full marks

1After n half-lives the fraction left is ( $\frac{1}{2}$ )ⁿ — list the halvings to avoid errors.
2Alpha decay: mass number -4, atomic number -2; beta decay: mass number same, atomic number +1.
3Half-life can be read off a graph as the time for the count rate to fall by half from any point.

Practice questions

Try these yourself

Start with the core skill, then open the answer only after you have attempted the full question.

1A sample falls from 640 counts per minute to 80 counts per minute in 12 hours. Find its half-life.

Mark scheme

1.Count the number of halvings.
2.Divide the total time by that number.

4 hours

2Explain why the activity never reaches exactly zero on a decay graph.

Mark scheme

1.Consider repeated halving.

Each half-life halves the remaining activity, so the curve approaches zero without reaching it exactly.

3A sample contains 48 mg of an isotope with a half-life of 5 years. Find the mass remaining after 20 years.

Mark scheme

1.Count four half-lives.
2.Halve four times.

3 mg

4Define the half-life of a radioactive isotope.[1 mark]

Mark scheme

1.Recall the standard definition involving halving.

The average time taken for the number of undecayed nuclei (or the activity) of a sample to halve (1)

5A sample has an activity of 800 Bq. Its half-life is 6 hours. Calculate its activity after 18 hours.[2 marks]

Mark scheme

1.Find the number of half-lives: 18 / 6.
2.Halve the activity that many times.

number of half-lives = 18 / 6 = 3 (1); 800 → 400 → 200 → 100, so activity = 100 Bq (1)

6Explain what is meant by saying radioactive decay is random.[2 marks]

Mark scheme

1.Consider predictability of single decays.

It is impossible to predict which nucleus will decay next (1) or exactly when a given nucleus will decay (1)

7A nucleus of mass number 226 and atomic number 88 emits an alpha particle. State the mass number and atomic number of the new nucleus.[2 marks]

Mark scheme

1.Alpha decay reduces mass number by 4.
2.Alpha decay reduces atomic number by 2.

mass number = 226 - 4 = 222 (1); atomic number = 88 - 2 = 86 (1)

8A radioactive source initially contains

6.4 \times 10^{6}

undecayed nuclei. After 20 days only

4.0 \times 10^{5}

nuclei remain. Calculate the half-life of the source.[3 marks]

Mark scheme

1.Find the fraction remaining: $4.0 \times 10^{5}$ / $6.4 \times 10^{6}$ .
2.Express as ( $\frac{1}{2}$ )ⁿ to find the number of half-lives.
3.Divide the total time by the number of half-lives.

fraction remaining =

4.0 \times 10^{5}

6.4 \times 10^{6}

\frac{1}{1}6

(1);

\frac{1}{1}6

= (

\frac{1}{2}

)⁴, so 4 half-lives have passed (1); half-life = 20 / 4 = 5 days (1)

9A nucleus of iodine-131 (atomic number 53) emits a beta particle. Write down the mass number and atomic number of the daughter nucleus and name the element produced (atomic number 54 is xenon).[3 marks]

Mark scheme

1.Beta decay: mass number unchanged, atomic number increases by 1.
2.New atomic number = 53 + 1 = 54.
3.New mass number = 131.
4.Element with atomic number 54 is xenon.

mass number = 131 (unchanged) (1); atomic number = 53 + 1 = 54 (1); the daughter nucleus is xenon-131 (1)

10A radioactive source gives a count rate of 320 counts per second (after correcting for background). 24 minutes later the count rate is 20 counts per second. Calculate the half-life of the source.[3 marks]

Mark scheme

1.Find the number of halvings: 320 → 160 → 80 → 40 → 20, so 4 halvings.
2.Number of half-lives = 4.
3.Half-life = 24 / 4.

320 / 20 = 16 = 2⁴, so 4 half-lives have elapsed (1); half-life = 24 / 4 (1) = 6 minutes (1)

11Explain why the activity of a radioactive source can never reach exactly zero, no matter how long you wait.[3 marks]

Mark scheme

1.After each half-life, activity halves.
2.Halving repeatedly gives smaller and smaller values.
3.Mathematically, ( $\frac{1}{2}$ )ⁿ approaches but never reaches zero.
4.In practice, very few nuclei remain but never exactly none.

After each half-life the activity is halved; after n half-lives the activity is (

\frac{1}{2}

)ⁿ of its original value (1); since halving a positive number always gives another positive number (never zero), the activity decreases but mathematically never reaches exactly zero (1); in practice, the number of nuclei becomes so small that statistically the source appears to have stopped, but this is because whole nuclei cannot be divided, not because activity truly reaches zero (1)

12Carbon-14 has a half-life of 5730 years and is used in radiocarbon dating. Explain how the technique works, and discuss one limitation of using carbon-14 dating for very old or very recent samples.[4 marks]

Mark scheme

1.Living organisms absorb C-14 from atmosphere; proportion stays constant.
2.At death no more C-14 absorbed; ratio of C-14 to C-12 falls.
3.Measure ratio and use half-life to calculate time since death.
4.Limitation for very old: too little C-14 left to measure accurately.
5.Limitation for recent: changes in atmospheric C-14 levels complicate calibration.

Living organisms continuously exchange carbon with the atmosphere, maintaining a constant ratio of radioactive carbon-14 to stable carbon-12 in their tissues; at death this exchange stops and the carbon-14 decays with a known half-life of 5730 years (1); by measuring the remaining ratio of carbon-14 to carbon-12 and comparing it to the ratio in living organisms, scientists can calculate how many half-lives have passed and therefore how long ago the organism died (1); for very old samples (over about 50 000 years) so few carbon-14 atoms remain that the count rate is too low to measure accurately, so the technique has a practical upper age limit (1); for very recent samples, changes in atmospheric carbon-14 levels caused by nuclear weapons testing and industrial burning of fossil fuels alter the baseline ratio, making accurate dating difficult without calibration (1)

13A nuclear equation for alpha decay is shown: Ra-226 (88) → Rn (86) + alpha particle. (a) Write down the mass number and atomic number of the radon nucleus. (b) The radon nucleus then undergoes beta decay. Write down the mass number and atomic number of the product nucleus. (c) Explain why, despite the decay sequence, the overall nuclear charge of the universe is conserved.[3 marks]

Mark scheme

1.(a) Alpha: mass number 226 - 4 = 222; atomic number 88 - 2 = 86.
2.(b) Beta: mass number unchanged = 222; atomic number 86 + 1 = 87.
3.(c) Charge is conserved in each step — both alpha and beta decays conserve total charge.

(a) Radon nucleus: mass number = 222, atomic number = 86 (1); (b) Beta decay product: mass number = 222, atomic number = 87 (1); (c) In alpha decay, the alpha particle (charge +2) and the daughter nucleus (charge 86) together have the same total charge as the parent (88); in beta decay, a neutron becomes a proton (charge +1) and an electron (charge -1), so no net charge is created; in each decay the total charge on both sides of the nuclear equation is equal, so overall nuclear charge is always conserved (1)

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