Posted in A2 Unit 5: Radioactivity, AQA A2 Unit 5 by Mr A on 11 Jan 2010

When atomic nuclei become unstable, they must emit either an α or β particle, or energy in the form of a γ ray. They become more stable in the process.

• A = activity of a sample, Bq (Becquerels) = the rate of decay = the number of nuclei that will decay in any given second
• N = the number of unstable nuclei remaining in a sample
• λ = the probability that any given nucleus will decay in any given second
• $T_{\frac{1}{2}}$ = the half-life of a sample, s (seconds)

If λ is the probability that any given nucleus will decay in a second, and there are N nuclei remaining, then it stands to reason that λN gives us the number of nuclei that will decay in a given second;

$A = \lambda N$

The number of nuclei remaining after some time Δt is reduced (hence the minus sign) by ΔN. Then, the activity of the sample is:

$A = -\frac{\Delta N}{\Delta t}$

Or,

$\frac{\Delta N}{\Delta t} = -\lambda N$

This is a linear differential equation whose solution is:

$N = N_{0} e^{- \lambda t}$

where

$N_{0}$ = the number of undecayed nuclei at time zero

N = the number of undecayed nuclei at time t

Thus, radioactive decay follows an exponential decay curve.

Now prove that

$T_{\frac{1}{2}} = \frac{ln 2}{\lambda}$

and

$A = A_{0} e^{- \lambda t}$