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6.3.2 Half Life

By the end of this lesson you can:

  • Define half-life
  • Calculate the activity or mass remaining after a given time
  • Interpret decay graphs
Find your course: NSW Year 11NSW Year 12: Module 8, IQ4QLD Unit 1: Ionising radiation and nuclear reactionsVIC Unit 1: Energy from the nucleusAP Physics 2: Unit 15IB: E.3 Radioactive decay
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🎥Watch
Learn the idea: a short explanation, then a check.

The Physics of Radioactive Half-Life

Radioactive decay is fundamentally both spontaneous (occurring without external influence) and random (the decay of an individual nucleus cannot be predicted). However, across a macroscopic sample containing a large number of nuclei (N), decay follows a predictable, statistical law.

1. The Radioactive Decay Law

The rate of decay, or activity (A), represents the number of disintegrations per unit time. It is directly proportional to the total number of undecayed radioactive nuclei (N) remaining in the sample:

A = - dN⁄dt = λN

Where λ is the constant of proportionality known as the decay constant (expressed in s-1 or s-1 equivalents), unique to each specific radioisotope. The negative sign indicates that the number of parent nuclei decreases over time.

2. Exponential Decay Formula

Integrating the rate equation yields the exponential decay function, which gives the remaining number of undecayed nuclei (N), mass (m), or activity (A) at any given time (t):

N(t) = N0 e-λt

Where N0 is the initial quantity of radioactive nuclei at t = 0.

3. Deriving Half-Life (t1/2)

The half-life (t1/2) is defined as the time required for half of the initial radioactive nuclei to decay. Setting N(t) = N0⁄2:

N0⁄2 = N0 e-λt1/2   &implies;   1⁄2 = e-λt1/2

Taking the natural logarithm (ln) of both sides gives the exact mathematical relationship between half-life and the decay constant:

t1/2 = ln(2)⁄λ ≈ 0.693⁄λ

4. Discrete Half-Life Calculation

For problem-solving over integer numbers of half-lives (n = t⁄t1/2), the remaining quantity can be calculated simplified as:

N(t) = N0 × (1⁄2)n
🕹️Explore
Test your predictions in a simulation.
Interactive
Lets' explore this concept further with this interactive from pHET


  1. Select Single Atom: Examine the decay of Polonium-211 into Lead-207 with the emission of an alpha particle.
  2. Reset and Repeat: Reset the nucleus multiple times after each decay. Notice that while individual decay events are completely random, the time taken for decay averages out to the element's half-life over many trials.
  3. Select Multiple Atoms: Add a large sample of atoms and run the simulation to observe decay behavior across a population.


Key Takeaways:

Although each individual atomic decay is random, predictable fractions decay over successive half-lives:


  • After 1 half-life: 50% decayed (50% remaining)
  • After 2 half-lives: 75% decayed (25% remaining)
  • After 3 half-lives: 87.5% decayed (12.5% remaining)
  • After 4 half-lives: 93.75% decayed (6.25% remaining)

This simulation does the same thing. Choose the isotope and set the time then start the simulation.

Experiment and discussion
What is the decay rate of a head of beer? This video looks at that and uses the results to help you understand the concept of half-life in radioactivity.
📝Practise
Worked examples first, then problems to try, then worksheets.
Sample Problem
We are now ready to try a sample problem
Below is a sample problem with a video that explain how to solve it. It is suggested you try the problem beforehand, as this actually aids understanding, even if you are unsure if you are correct.
Picture

Practice Problems: Half-Life

Half-lives used: arsenic-81 (As-81) = 33 s; gold-198 (Au-198) = 2.69 days; carbon-14 (C-14) = 5,730 years. After n half-lives, the fraction remaining is (½)n.

1. How long does it take a 100.00 g sample of arsenic-81 to decay to 6.25 g?

Given: N0 = 100.00 g, N = 6.25 g, T½ (As-81) = 33 s

Step 1 (Number of half-lives): N0 / N = 100.00 / 6.25 = 16 = 24, so n = 4 half-lives. (Check: 100 → 50 → 25 → 12.5 → 6.25.)

Step 2 (Time): t = n × T½ = 4 × 33 s = 132 s

Answer: 132 s
2. What percentage of an arsenic-81 sample remains undecayed after 43.2 seconds?

Given: t = 43.2 s, T½ (As-81) = 33 s

Step 1 (Number of half-lives): n = t / T½ = 43.2 / 33 = 1.309

Step 2 (Fraction remaining): N / N0 = (½)n = (0.5)1.309 = 0.4036

Step 3 (Percentage): 0.4036 × 100% ≈ 40.4%

Answer: 40.4%
3. How old is an ancient bone if it currently contains 0.3125 g of carbon-14, but originally contained 80.000 g of carbon-14?

Given: N0 = 80.000 g, N = 0.3125 g, T½ (C-14) = 5,730 years

Step 1 (Number of half-lives): N0 / N = 80.000 / 0.3125 = 256 = 28, so n = 8 half-lives.

Step 2 (Age): t = n × T½ = 8 × 5,730 years = 45,840 years

Answer: 45,840 years
🍎For teachers
Ideas for using this lesson in class, and where to find the answers.
If you wish to the experiment yourself , here is the experiment on half life