Radioactive decay and half-lifeEdexcel International A Level Physics: Revision notes

Nuclear decay is spontaneous: it is not affected by external factors such as temperature, pressure or chemical state, and nothing needs to trigger it.

It is also random: it is impossible to predict when a particular nucleus will decay. Every nucleus has the same constant probability of decaying in each unit of time. When a large number of nuclei is present, the number that decay in a fixed time is predictable on average, but individual counts fluctuate about a mean value.

This is why successive counts from a steady source, such as 52, 47, 58 and 44 in 10 s intervals, are not identical.

The activity $A$ of a sample is the number of nuclei that decay per second, measured in becquerels (Bq). It is proportional to the number $N$ of undecayed nuclei:

$A = \lambda N$

where the decay constant $\lambda$ is the probability of decay of a nucleus per unit time. Its unit is s⁻¹. Because the number of nuclei falls as they decay:

$\dfrac{dN}{dt} = -\lambda N$

The minus sign shows that $N$ is decreasing. A large $\lambda$ means a fast decay and a short half-life.

Solving $dN/dt = -\lambda N$ gives the equation for the number of nuclei:

$N = N_0 e^{-\lambda t}$

Because $A = \lambda N$, the activity (and the corrected count rate) follows the same law:

$A = A_0 e^{-\lambda t}$

Taking natural logarithms gives the log equation, which is a straight line:

$\ln N = \ln N_0 - \lambda t \qquad \ln A = \ln A_0 - \lambda t$

So a graph of $\ln A$ against $t$ has a gradient of $-\lambda$ and an intercept of $\ln A_0$.

The half-life $t_{1/2}$ is the time taken for the number of undecayed nuclei, or the activity, to fall to half its value. Put $N = N_0/2$ at $t = t_{1/2}$ in $N = N_0 e^{-\lambda t}$:

$\tfrac{1}{2} = e^{-\lambda t_{1/2}} \;\Rightarrow\; \ln 2 = \lambda t_{1/2} \;\Rightarrow\; \lambda = \dfrac{\ln 2}{t_{1/2}}$

The half-life depends only on $\lambda$, so it is the same whatever the size of the sample: the value halves every half-life, however many nuclei there are.

Worked example: iodine-131 has $t_{1/2} = 8.0$ days $= 6.91 \times 10^5$ s, so $\lambda = 0.693 \div 6.91 \times 10^5 = 1.00 \times 10^{-6}$ s⁻¹. A sample with $A = 6.4 \times 10^5$ Bq has $N = A/\lambda = 6.4 \times 10^{11}$ nuclei.

From an activity-time graph: read the time for the activity to fall from some value to half of it. Check this with a second halving, e.g. from 480 Bq to 240 Bq, then from 240 Bq to 120 Bq. If the times are equal, the decay is exponential and the half-life is that time.

From a log graph: plot $\ln A$ against $t$. The line is straight with gradient $-\lambda$, so

$\lambda = -\text{gradient}$ and $t_{1/2} = \ln 2 \div \lambda$

For example, between 0 s and 160 s the activity falls from 480 Bq to 120 Bq, so the gradient is $(\ln 120 - \ln 480) \div 160 = -8.7 \times 10^{-3}$ s⁻¹, so $t_{1/2} = 0.693 \div 8.7 \times 10^{-3} = 80$ s.

Remember that a real count rate must be corrected for background before it is used.