Question

(a) Derive the mathematical expression for law of radioactive decay for a sample of a radioactive nucleus.
(b) How is the mean life of a given radioactive nucleus related to the decay constant?

Answer 1

Radioactive decay of a radioactive substance is directly proportional to no of nuclei of the radioactive atoms use this property to find the mathematical expression for the law of radioactive decay and once you know the expression you can also find the relationship between mean life and decay constant.

Complete step by step answer:
We know that for radioactive materials rate at which nuclei depends is directly proportional to no of nuclei present we do not know for sure that which nuclei will decay in a given time period but no of nuclei decayed in a given time interval is always proportional to the no of nuclei

Therefore,
$\dfrac{{dN}}{{dt}} \propto N$
$\Rightarrow \dfrac{{dN}}{{dt}} = - \lambda N$
Where $\lambda$ is decay constant and minus sign is given to imply that number of nuclei decreases over time.

$\Rightarrow \dfrac{{dN}}{N} = - \lambda dt$
Integrating both sides we have,
$\int {\dfrac{{dN}}{N} = \int { - \lambda dt} } \\\ \Rightarrow \ln N = - \lambda t + C$
At t=0 value of N is ${N_0}$
$\Rightarrow \ln \dfrac{N}{{{N_0}}} = - \lambda t$
Or $N = {N_0}{e^{ - \lambda t}}$
Lets call rate of decay as R then ,
$R = \dfrac{{ - dN}}{{dt}} \Rightarrow R = - \lambda N \Rightarrow R = - \lambda {N_0}{e^{ - \lambda t}}$

Now we will find the value of the mean life of the radioactive decay. Let's find the relation between the mean life $\tau$ and the disintegration constant λ. The number of nuclei which decay in the time interval: ‘t’ to ‘t + dt’ is:
$Rdt = \lambda {N_0}{e^{ - \lambda t}}dt$
Where dt is infinitesimal time interval

Each of them has lived for time ‘t’. Hence, the total life of all these nuclei is $t\lambda {N_0}{e^{ - \lambda t}}dt$.
Hence, to obtain the mean life, we integrate this expression over all the times from 0 to $\infty$ and divide by the total number of nuclei at t = 0 (which is ${N_0}$).
$\tau = \lambda {N_0}\int\limits_0^\infty {t{e^{ - \lambda t}}dt}$
Where $\tau$ is mean life,
On solving this integral, we get
$\tau = \dfrac{1}{\lambda }$

Thus correct answer for (a) is $N = {N_0}{e^{ - \lambda t}}$ and for (b) is $\tau = \dfrac{1}{\lambda }$.

Note: Use the law of radioactive decay to find the expression for no of nuclei at any time once you know the expression for number of nuclei at any time simply then find the value of half life and average life using the mathematical formula of half life and average life.