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NotesPhysics HLTopic 5.3Half-life, activity and background radiation
Back to Physics HL Topics
5.3.42 min read

Half-life, activity and background radiation

IB Physics • Unit 5

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Contents

  • Half-life: the count rate halves
  • Counting over whole half-lives
  • Exam-style question
The big idea: Radioactive decay is random — you can't say when one nucleus will decay. But on average a sample gets weaker with time.

The half-life is the time it takes for half the radioactive nuclei to decay.

In that same time the activity and the count rate also halve.
New words, plainly: Activity = how many nuclei decay each second, measured in becquerel (Bq), where 1 Bq = 1 decay per second.

Count rate = how many of those decays a detector actually records each second (clicks per second).

Half-life = the time for the activity (or count rate) to fall to half.
Halving keeps going: Every half-life that passes, you halve again — not subtract a fixed amount.

So after 1, 2, 3 half-lives the count rate is 1/2, 1/4, 1/8 of the start. It flattens but never reaches zero.

At SL you only deal with whole numbers of half-lives, so you never need the exponential equation. You just halve, halve, halve, once for each half-life that has passed.

The rule (learn this — it is NOT in the data booklet): After n whole half-lives, the count rate (or activity) is the start value multiplied by (1/2)ⁿ.

Work out n = total time ÷ half-life, then halve that many times.
Half-lives passedFraction of original leftCount rate from 80 s⁻¹
01 (all of it)80 s⁻¹
11/240 s⁻¹
21/420 s⁻¹
31/810 s⁻¹
n(1/2)ⁿ80 × (1/2)ⁿ
Subtract the background FIRST: A detector always records some background radiation (from rocks, cosmic rays, etc.) even with no source.

The true count rate from the source = measured count rate − background count rate. Correct for background before you start halving.

Worked example — count rate after two half-lives

A fresh source gives a measured count rate of 204 counts per second. The background count rate is 4 counts per second. The source has a half-life of 6.0 hours. Find the MEASURED count rate 12 hours later.

Solution

  1. Correct for background first. True count rate from the source now:
  2. Find how many half-lives have passed: n = total time ÷ half-life.
  3. Halve the source count rate twice (once per half-life):
  4. Add the background back on — the detector still records it:

Final answer

The measured count rate after 12 hours is 54 counts per second.

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How this is tested: Half-life is a classic Paper 1A (multiple-choice) one-step calculation, and turns up in Paper 2 structured questions too.

- Paper 1A — halve over whole half-lives: given a start value and a half-life, find the count rate / activity after a whole number of half-lives. - Paper 1A — correct for background: subtract the background count before halving (and add it back if asked for the measured value). - Paper 1A — ratios: two samples with the same half-life keep the same ratio of activities as both halve together. - Paper 2: use the activity to work out an effect, such as an ionisation current.

Classic trap: forgetting the background, or subtracting a fixed amount each half-life instead of halving.
Ratios stay put: If two sources have the same half-life, after the same time both have halved the same number of times.

The ratio of their activities is unchanged — e.g. 5 : 1 stays 5 : 1.

IB-style question — (a) activity after three half-lives

A radioactive sample has an initial activity of 6.4 × 10⁶ Bq and a half-life of 8.0 days. Determine its activity after 24 days.

Solution

  1. Find the number of half-lives: n = total time ÷ half-life.
  2. Halve the activity three times (once per half-life):
  3. Tidy the answer with its unit (Bq = decays per second):

Final answer

After 24 days (3 half-lives) the activity is 8.0 × 10⁵ Bq — one-eighth of the start.

IB-style question — (b) ratio of two samples

Two samples X and Y have the SAME half-life. Right now sample X is four times as active as sample Y. State the ratio of their activities after two half-lives, and explain your answer.

Solution

  1. Both samples have the same half-life, so in the same time both halve the same number of times.
  2. After two half-lives each activity is multiplied by (1/2)² = 1/4 — the same factor for both.
  3. Multiplying both by the same factor leaves their ratio unchanged.

Final answer

The ratio stays 4 : 1. Equal half-lives means both fall by the same factor, so the ratio is unchanged.

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what is meant by the activity of a radioactive source, and its SI unit. [2 marks]

Related Physics HL Topics

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5.1.1Nuclear model and atomic structure
5.1.2Energy levels and atomic spectra
5.1.3The electronvolt
5.1.4Quantisation of charge
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5.3.3Energy released in radioactive decay
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