Radioactive decay
Random, yet predictable
- You can never say when a single nucleus will decay — it is random.
- Yet a huge sample of them decays in a smooth, predictable curve.
- Probability turns chaos into a reliable law.
Random and spontaneous
- Decay is spontaneous — no trigger; unaffected by temperature, pressure or chemistry.
- It is random — a Geiger counter clicks at uneven intervals, never a steady stream.
Select all the true statements about radioactive decay.
It is spontaneous and random, and the rate does not depend on conditions. You can only give the probability of a single decay.
Activity and decay constant
- Activity $A = \lambda N$ — decays per second (unit: becquerel, Bq).
- $\lambda$ is the decay constant — the probability per second that a nucleus decays.
The activity of a radioactive source is:
Activity = decay constant × number of undecayed nuclei, measured in becquerel.
A source has $\lambda = 0.010$ per second and $1.0 \times 10^{6}$ undecayed nuclei. What is its activity?
$A = \lambda N = 0.010 \times 1.0 \times 10^{6} = 1.0 \times 10^{4}\ \text{Bq}$.
Exponential decay
- The same fraction decays each second, so $N = N_0 e^{-\lambda t}$ (and $A = A_0 e^{-\lambda t}$).
- A $\ln$ plot is a straight line of gradient $-\lambda$.
The activity of a radioactive source decays exponentially with time.
The same fraction decays each second, giving $A = A_0 e^{-\lambda t}$.
Half-life
- The half-life $t_{1/2}$ is the time to fall to half: $t_{1/2} = \dfrac{\ln 2}{\lambda}$.
- After $n$ half-lives, a fraction $\left(\tfrac{1}{2}\right)^{n}$ remains.
The half-life is related to the decay constant by:
Setting $N = \tfrac{1}{2}N_0$ in $N = N_0 e^{-\lambda t}$ gives $\lambda t_{1/2} = \ln 2$.
What fraction of the nuclei remain after 3 half-lives?
Each half-life halves the number: $\left(\tfrac{1}{2}\right)^{3} = \tfrac{1}{8} = 0.125$.
You've got it
- decay is spontaneous and random (rate fixed, timing unpredictable)
- activity $A = \lambda N$ (Bq); $N = N_0 e^{-\lambda t}$
- half-life $t_{1/2} = \dfrac{\ln 2}{\lambda}$