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

Radioactive decay is spontaneous: it is not triggered by an external cause, and it is unaffected by temperature, pressure or chemical state.

It is also random: every nucleus has the same fixed probability of decaying in a given time, but it is impossible to predict when a particular nucleus will decay. In a large sample the number of decays in a time interval varies about an average value, so repeated counts of a constant source differ slightly. The percentage fluctuation is smaller when the count is larger, so count for longer or repeat and average.

The activity $A$ of a sample is the number of nuclei that decay per second, measured in becquerels (Bq, 1 Bq = 1 decay per second).

The activity is proportional to the number of undecayed nuclei $N$:

$A = \lambda N \qquad \frac{dN}{dt} = -\lambda N$

The constant $\lambda$ is the decay constant, the probability of decay per nucleus per second, with unit s⁻¹. The minus sign shows that $N$ decreases. As nuclei decay, $N$ falls, so the activity falls too.

Worked example: $N = 2.4 \times 10^{15}$ and $\lambda = 1.0 \times 10^{-6}$ s⁻¹, so $A = \lambda N = 2.4 \times 10^{9}$ Bq.

The solution of $dN/dt = -\lambda N$ is

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

and because $A \propto N$ (and the corrected count rate $\propto A$),

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

The half-life $t_{1/2}$ is the time for the number of undecayed nuclei, or the activity, to fall to half its initial value. Putting $N = N_0/2$ gives

$t_{1/2} = \frac{\ln 2}{\lambda}$

After $n$ half-lives the activity is $A_0/2^n$.

Worked example: half-life 6.0 h, so $\lambda = 0.693/(6.0 \times 3600) = 3.2 \times 10^{-5}$ s⁻¹. A source falls from 400 to 60 MBq when $\ln(400/60) = \lambda t$, so $t = 1.90/(3.2 \times 10^{-5}) = 5.9 \times 10^{4}$ s, or 16 hours.

Taking natural logarithms of $A = A_0 e^{-\lambda t}$ gives

$\ln A = \ln A_0 - \lambda t$

so a graph of $\ln A$ (or $\ln N$) against $t$ is a straight line with gradient $-\lambda$ and intercept $\ln A_0$. This gives a more reliable value than reading single points off a curve.

Rearranging for time: $t = \dfrac{1}{\lambda}\ln\dfrac{A_0}{A}$.

From the decay curve. Plot corrected count rate against time. Read the time for the rate to fall from any value to half of it, and repeat from different starting values. If the half-life is constant, the decay is exponential. Averaging several values reduces error.

From the log graph. Plot $\ln A$ against $t$, draw a best-fit line through all the points and find its gradient. Then $\lambda = -\text{gradient}$ and $t_{1/2} = \ln 2/\lambda$.

Always use the corrected count rate (background subtracted). At low count rates the random fluctuations are a larger fraction of each reading, so the late points have larger percentage uncertainty.

Worked example: a count rate falls 800 to 400 in 20 minutes, so $t_{1/2} = 20$ min and $\lambda = 0.0347$ min⁻¹. After 90 minutes $A = 800\,e^{-0.0347 \times 90} = 35$ counts per minute.