K223 / K224 / 2027
Radioactivity overview

Lesson 4 of 5

Half-life from numbers and graphs

  • I can use half-life in simple calculations, tables and decay curves.Syllabus

    Syllabus K223 / K224, 13(g). Use half-life in simple calculations, tables and decay curves.

Make a guess. You are not marked.

A source has a half-life of 10 minutes. After 20 minutes, what fraction of the original number of undecayed nuclei is left?

  1. One half, because only one more half-life has passed since the first.
  2. None. Two half-lives make a whole.
  3. One quarter, after halving twice.
Show the answer

One quarter, after halving twice.

Twenty minutes is two half-lives. After the first, half remains. After the second, half of that half remains: one quarter.

In short

  • Half-life is the time for half the radioactive nuclei in a sample to decay.
  • After each half-life, the corrected count rate halves.
  • After two half-lives, one quarter remains.

Half-life is the time taken for half the undecayed radioactive nuclei in a large sample of one nuclide to decay. Over that time, its activity falls to half.

Use the background-corrected count rate. If the detector stays in the same place with the same settings, the corrected count rate halves whenever the activity halves.

The symbol is t1/2. It is a time, with SI unit s. A problem may use minutes, hours or days consistently; compare the elapsed time and half-life in matching units.

Halve what remains each time

Repeated halving

A corrected count rate of 800 counts/min, with a half-life of 6 h

  • After 6 h: 800/2 = 400 counts/min.
  • After 12 h: 400/2 = 200 counts/min.
  • After 18 h: 200/2 = 100 counts/min.

Eighteen hours contains three half-lives. The remaining fraction is (1/2) x (1/2) x (1/2) = 1/8; the decayed fraction is 7/8. The same 6 h half-life is 6 x 3600 = 21600 s.

Try it step by step

Work it out in steps: Number of half-lives, then Corrected count rate remaining in counts/min.

  1. Problem 1. A source gives a corrected count rate of 1200 counts/min. Its half-life is 8 minutes. What will the corrected count rate be after 24 minutes?
  2. Problem 2. A source gives a corrected count rate of 640 counts/min. Its half-life is 5 hours. What will the corrected count rate be after 20 hours?
  3. Problem 3. A source gives a corrected count rate of 2000 counts/min. Its half-life is 3 days. What will the corrected count rate be after 12 days?
Show the answer

Problem 1.

Number of half-lives: 24 / 8 = 3 half-lives

Corrected count rate remaining in counts/min: 1200 -> 600 -> 300 -> 150 counts/min

Problem 2.

Number of half-lives: 20 / 5 = 4 half-lives

Corrected count rate remaining in counts/min: 640 -> 320 -> 160 -> 80 -> 40 counts/min

Problem 3.

Number of half-lives: 12 / 3 = 4 half-lives

Corrected count rate remaining in counts/min: 2000 -> 1000 -> 500 -> 250 -> 125 counts/min

Half of the radioactive parent nuclei remain after one half-life on average. Half of the whole sample has not vanished: daughter material is still present. After the next half-life, half of the remaining parents decay.

Find half-life from corrected data

This separate supplied model follows one radioactive nuclide. Background has a constant mean of 20 counts/min; daughter contributions are neglected. The rates represent the labelled instants, with detection conditions unchanged.

Subtract the same mean background from each total rate.
Time / minTotal / (counts/min)Source / (counts/min)
0180160
410080
86040
124020
163010

Remove the background before finding the half-life

Supplied smooth model: one radioactive nuclide, unchanged detection conditions, constant mean background of 20 counts/min and negligible daughter contribution. Real readings fluctuate around such a trend.

Total count rate includes the background

Total count rate includes the backgroundAn ideal smooth total-count-rate curve uses time in minutes horizontally and counts per minute vertically. Marked values are 180 at zero minutes, 100 at four, 60 at eight, 40 at twelve, and 30 at sixteen. The dashed horizontal line is the constant mean background of twenty counts per minute. The curve remains above that line and approaches it as the source contribution becomes smaller. It does not approach zero. These are supplied model values, not invented experimental measurements.Count rate / counts per min02040801201602000481216Time / min

The dashed line is the 20 counts/min background. The total curve approaches that level, rather than zero.

Net source rate: subtract 20 counts/min

Net source rate: subtract 20 counts/minOn the same axes and scales, the corrected smooth source-rate curve passes through 160, 80, 40, 20 and 10 counts per minute at zero, four, eight, twelve and sixteen minutes. Two shaded intervals, zero to four minutes and eight to twelve minutes, each last four minutes. In the first, 160 halves to 80; in the second, 40 halves to 20. Brackets below the graph label the equal durations. The corrected curve approaches zero but is still ten counts per minute at sixteen minutes. It is a continuous ideal model for one nuclide with unchanged detection conditions and no daughter contribution.Count rate / counts per min02040801201602000481216Time / min4 min160 to 804 min40 to 20

160 to 80 and 40 to 20 counts/min each take 4 min. The net rate approaches zero; it is still 10 counts/min at 16 min.

The axes use the same scales in both panels. Each marked dot matches a supplied table value; the curve represents the ideal average pattern between them.

The total-rate curve approaches the background level of 20 counts/min. The corrected source curve approaches zero. The marked 0-4 min and 8-12 min intervals both halve the source contribution.
  1. Correct the first pair: 180 - 20 = 160 and 100 - 20 = 80 counts/min.
  2. Read the time interval: 160 to 80 takes 4 - 0 = 4 min.
  3. Check another pair: the net rate falls from 40 at 8 min to 20 at 12 min, again taking 4 min.

A fall from 160 to 20 counts/min is three halvings: 160, 80, 40, 20. It therefore takes 3 x 4 = 12 min. It is not one half-life simply because the final reading is small.

Halving the uncorrected 180 to 90 counts/min would also halve the background contribution. The background does not follow the source's decay, so that would give the wrong interval.

Read a curve and interpret real measurements

Label elapsed time horizontally and the stated rate vertically. On a corrected curve, choose a source rate, find half that rate, and read the difference between their times. Use another well-separated pair as a check.

The source loses the same fraction over equal half-life intervals, not the same amount. The average curve bends as the rate falls; it does not become zero after two half-lives. Individual measurements scatter around this trend because decay and detection are random.

For a logged count series, retain the timestamps and counting durations. Calculate rates, subtract a comparable background estimate and keep source-detector geometry fixed. A short counting window can give an approximate rate at its labelled time, but an excessively long window blurs a rapid decay. Do not mistake a changed detector position or setting for a new half-life.

Watch out for this

A student says: "After two half-lives, none of the radioactive nuclei are left." What is wrong with this?

Show the answer

Each half-life halves the number that remain, not the original number. After two half-lives, one quarter of the original nuclei are still undecayed.

Graph reader

Find a half-life from measured counts

A detector records the total count rate from a source plus background. Move the time cursor to read the graph.

050100150200250300051015202530time / mincount rate / (counts/min)

Check your understanding

A single-nuclide model has background 20 counts/min. Its total rates are 180 counts/min at 0 min, 100 at 4 min and 60 at 8 min. What half-life follows from these readings?

  1. It cannot be found until the total rate reaches zero.
  2. 4 min: the corrected rate falls from 160 to 80, then from 80 to 40 counts/min.
  3. Between 4 and 8 min, when the total rate falls from 180 to 90 counts/min.
Show the answer

4 min: the corrected rate falls from 160 to 80, then from 80 to 40 counts/min.

Both corrected halving intervals last 4 min under unchanged detection conditions. Equal time intervals halve the remaining source contribution; they do not subtract an equal number of counts each time.

Practise exam questions

Write your answer first, then tick the mark points it makes.

Exam-style question

A sample contains 80 g of a radioactive isotope with a half-life of 6.0 days. Calculate the mass of this isotope that remains undecayed after 18 days. [2]

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Exam-style question

Define half-life. [1]

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