9478 / 2027
Nuclear Physics overview

Lesson 4 of 10

Half-life and corrected curves

  • I can infer and sketch exponential decay and use x = x0 exp(-lambda t) for an appropriate population, activity or received source rate.Syllabus

    Syllabus 9478, 20(i). Infer and sketch exponential decay and use x = x0 exp(-lambda t) for an appropriate population, activity or received source rate.

  • I can define half-life and solve problems using suitable quantities, tables or decay curves.Syllabus

    Syllabus 9478, 20(j). Define half-life and solve problems using suitable quantities, tables or decay curves.

Make a guess. You are not marked.

Two samples contain the same radioactive nuclide. Sample B has twice as many undecayed nuclei as sample A. How do their half-lives and activities compare?

  1. Same half-life. B has twice the activity.
  2. B has twice the half-life and twice the activity.
  3. Same half-life and the same activity.
Show the answer

Same half-life. B has twice the activity.

Half-life is a property of the nuclide, not of the amount. Activity is lambda times the number of nuclei, so twice the nuclei means twice the decays per second.

Half-life is the time for the expected number of undecayed parent nuclei, or the activity, to fall to half its initial value. For a detector reading, use the source contribution and check that the conditions make it proportional to the decaying population.

For a single unchanged radionuclide, the expected number of undecayed parent nuclei and the activity halve over the same interval. A proportional net source count rate does too when detection conditions remain fixed. Half-life is not the lifetime of every individual nucleus.

After one, two and three half-lives, the remaining fractions are 1/2, 1/4 and 1/8. A few half-lives leave a fraction, not zero. Match time units before comparing: 0.2 min is 12 s, while 2 min is 120 s.

Separate total rate from the source trend

In this supplied model, net source rate starts at 80.0 counts/s and halves every 120 s. Background remains 4.0 counts/s. These are instantaneous expected rates at the named times, not long-window measured averages.

Expected net and total count rates in the 120 s half-life model
t / sNet / counts s-1Total / counts s-1
08084
1204044
2402024
3601014
48059

Total expected rate approaches the background

Total expected rate approaches the backgroundBoth decay panels use identical time and rate scales, over zero to 480 seconds. The expected source rate begins at 80 counts per second and halves every 120 seconds. Adding a constant background of four counts per second gives total values 84, 44, 24, 14 and 9 at successive 120-second times. The curve tends towards the dashed background line; it never becomes that value at a finite time. The marked values are ideal instantaneous model rates, not raw counts or long-window averages.Total rate / counts s-10204060800120240360480t / sBackground: 4

The same fixed background is added at every time. Half of the initial raw total is 42 counts/s; that is not the correct halving target for the source. At 120 s the total is 44 counts/s.

Corrected expected rate halves in equal times

Corrected expected rate halves in equal timesBoth decay panels use identical time and rate scales, over zero to 480 seconds. The expected source rate begins at 80 counts per second and halves every 120 seconds. Subtracting the background gives 80, 40, 20, 10 and 5 counts per second at successive 120-second times. Guides identify 80 to 40 and 40 to 20 with two adjacent 120-second brackets. The smooth expected net rate tends towards zero, not to a negative activity and not to zero at a finite plotted time. The marked values are ideal instantaneous model rates, not raw counts or long-window averages.Net rate / counts s-10204060800120240360480t / s120 s120 s

The two marked intervals each last 120 s even though the absolute decrease is smaller in the second interval. A proportional corrected rate shares the half-life only while source and detection conditions remain fixed.

Both expected-rate plots use the same time and rate scales. Total approaches the 4 counts/s background; net approaches zero. Equal 120 s spans halve the corrected rate from different starting points.

Half of the initial raw total is 42 counts/s, but the total at one half-life is 44 counts/s: 40 from the source plus 4 background. Read half-life from the corrected quantity. The falling trend becomes less steep as fewer parents remain; it is not a constant rate of loss.

Worked transfer problem

Count the halvings

A corrected rate decreases from 72 to 9 counts/s in 15 min under fixed detection conditions:

72 → 36 → 18 → 9
Three half-lives = 15 min
Half-life = 5.0 min

One eighth of the initial rate represents three halvings, not eight. A small negative background-subtracted estimate near zero can arise from fluctuations; it is not negative physical activity.

Try it step by step

Work it out in steps: Number of half-lives, then Fraction of nuclei remaining, then Activity remaining in Bq.

  1. Problem 1. A source has an activity of 1200 Bq and a half-life of 8.0 hours. Find its activity after 24 hours.
  2. Problem 2. A source has an activity of 5000 Bq and a half-life of 15 minutes. Find its activity after 60 minutes.
  3. Problem 3. A source has an activity of 8000 Bq and a half-life of 2.0 days. Find its activity after 10 days.
Show the answer

Problem 1.

Number of half-lives: 24 / 8.0 = 3 half-lives

Fraction of nuclei remaining: Fraction = (1/2)3 = 0.125

Activity remaining in Bq: A = 1200 x 0.125 = 150 Bq

Problem 2.

Number of half-lives: 60 / 15 = 4 half-lives

Fraction of nuclei remaining: Fraction = (1/2)4 = 0.0625

Activity remaining in Bq: A = 5000 x 0.0625 = 313 Bq (3 sf)

Problem 3.

Number of half-lives: 10 / 2.0 = 5 half-lives

Fraction of nuclei remaining: Fraction = (1/2)5 = 0.03125

Activity remaining in Bq: A = 8000 x 0.03125 = 250 Bq

The exponential model relates this shape to decay constant, activity and a logarithmic data fit.

Try it step by step

Work it out in steps: Decay constant lambda = ln 2 / half-life in per hour, then Activity A = A0 e^(-lambda t) in Bq.

  1. Problem 1. A source has a half-life of 10 hours and an initial activity of 2000 Bq. Find its activity after 5.0 hours.
  2. Problem 2. A source has a half-life of 6.0 hours and an initial activity of 900 Bq. Find its activity after 4.0 hours.
  3. Problem 3. A source has a half-life of 20 hours and an initial activity of 1500 Bq. Find its activity after 30 hours.
Show the answer

Problem 1.

Decay constant lambda = ln 2 / half-life in per hour: lambda = 0.693 / 10 = 0.0693 per hour

Activity A = A0 e^(-lambda t) in Bq: A = 2000 x e^(-0.0693 x 5.0) = 2000 x 0.707 = 1410 Bq (3 sf)

Problem 2.

Decay constant lambda = ln 2 / half-life in per hour: lambda = 0.693 / 6.0 = 0.116 per hour

Activity A = A0 e^(-lambda t) in Bq: A = 900 x e^(-0.116 x 4.0) = 900 x 0.630 = 567 Bq (3 sf)

Problem 3.

Decay constant lambda = ln 2 / half-life in per hour: lambda = 0.693 / 20 = 0.0347 per hour

Activity A = A0 e^(-lambda t) in Bq: A = 1500 x e^(-0.0347 x 30) = 1500 x 0.354 = 530 Bq (3 sf)

Check your understanding

With fixed detection conditions, a background-corrected rate falls from 72 to 9 counts/s in 15 min. What is the half-life?

  1. 7.5 min.
  2. 1.875 min.
  3. 5.0 min.
Show the answer

5.0 min.

72 to 36 to 18 to 9 is three half-lives. Therefore each half-life is 15/3 = 5.0 min.