Question

(A) Define ‘activity’ of a radioactive substance.
(B) Two different radioactive elements with half-lives ${T_1}$ and ${T_2}$ have ${N_1}$ and ${N_2}$ undecayed atoms respectively present at a given instant. Derive an expression for the ratio of their activities at this instant in terms of ${N_1}$ and ${N_2}$.

Answer 1

Hint: We will use the relation $R = - \dfrac{{dN}}{{dt}}$, to define the total decay rate $R$ of a substance of radionuclides. Also, the relation for the leftover number of undecayed atoms is given as $N = {N_0}{e^{ - \lambda t}}$, where ${N_0}$ is the number of initial atoms and $\lambda $ is a constant.

Complete step by step answer:
The ‘activity’ in radioactive decay processes is the number of disintegrations per second, i.e. the number of unstable atomic nuclei that decay per second in a given sample. This ‘activity’ is determined using radiation detectors and electronic circuits. The total decay rate $R$ of a substance of radionuclides is also called the ‘activity’ of that sample. The activity, or total decay rate $R$ is given by $R = - \dfrac{{dN}}{{dt}}$, where $N$ is the number of undecayed atoms. This leftover number of undecayed atoms is given as $N = {N_0}{e^{ - \lambda t}}$, where ${N_0}$ is the number of initial atoms.
Therefore we find $R$ from the above equations as
$R = - \dfrac{{dN}}{{dt}} = - \dfrac{{d({N_0}{e^{ - \lambda t}})}}{{dt}} = - \lambda ({N_0}{e^{ - \lambda t}}) = - \lambda N$
At the half-life of a substance, the number of left-over atoms reduces to half of its initial value. Thus, using the formula $N = {N_0}{e^{ - \lambda t}}$ at the half-life ${T_{1/2}}$, we get,
$\dfrac{1}{2}{N_0} = {N_0}{e^{ - \lambda {T_{1/2}}}}$
Taking the natural log on both sides, we get,
$\ln (\dfrac{1}{2}) = \ln ({e^{ - \lambda {T_{1/2}}}})$
Upon further solving we get,
$\ln 2 = \lambda {T_{1/2}}$,
$ \Rightarrow \lambda = \dfrac{{\ln 2}}{{{T_{1/2}}}}$
We will use these formulae and find the relation between $N$ and the half-life ${T_{1/2}}$.
Substituting $\lambda = \dfrac{{\ln 2}}{{{T_{1/2}}}}$ in $R = - \lambda N$, we get
$R = - \dfrac{{\ln 2}}{{{T_{1/2}}}}N$.
Now, for two different radioactive elements, with half-lives ${T_1}$ and ${T_2}$ have ${N_1}$ and ${N_2}$ undecayed atoms respectively present at a given instant, we have
${R_1} = - \dfrac{{\ln 2}}{{{T_1}}}{N_1}$and ${R_2} = - \dfrac{{\ln 2}}{{{T_2}}}{N_2}$.
Eliminating $\ln 2$ by dividing both the equations, we get, $\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{T_2}}}{{{T_1}}}\dfrac{{{N_1}}}{{{N_2}}}$. This is the required ratio of their activities.

Note: From the result above, it is clear that the ratio of ‘activity’ of the different radioactive elements with half-lives ${T_1}$ and ${T_2}$, and having ${N_1}$ and ${N_2}$ undecayed atoms respectively is dependent only on the number of leftover radioactive atoms $N$ and the half-life ${T_{1/2}}$.