Question

If the decay constant of a radioactive substance is $\lambda$ , then its half -life and mean life are respectively
(A) $\dfrac{1}{\lambda }$and $\dfrac{{{{\log }_e}2}}{\lambda }$
(B) $\dfrac{{{{\log }_e}2}}{\lambda }$and $\dfrac{1}{\lambda }$
(C) $\lambda {\log _e}2$and $\dfrac{1}{\lambda }$
(D) $\dfrac{\lambda }{{{{\log }_e}2}}$and $\dfrac{1}{\lambda }$

Answer 1

Hint The rate of radioactive decay is directly proportional to the concentration of nuclei of radioactive substance. And the proportionality constant is called the decay constant. We will use the law of radioactive decay to solve this problem.

Complete step-by-step solution:
The law of radioactive decay formulated as,
$N\left( t \right) = {N_0}{e^{ - \lambda t}}$
Where, N(t) is the number of radioactive nuclei remaining at time t
${N_0}$ is the initial number of radioactive nuclei And, $\lambda$ is the decay constant.
Now, the half life is the time after which the number of radioactive nuclei remains half of the initial number of nuclei, that is, when $N\left( t \right)$ becomes half of ${N_0}$.
So, $\dfrac{{{N_0}}}{2} = {N_0}{e^{ - \lambda {t_{\dfrac{1}{2}}}}}$
${e^{ - \lambda {t_{\dfrac{1}{2}}}}} = \dfrac{1}{2}$
$\lambda {t_{\dfrac{1}{2}}} = - {\log _e}\dfrac{1}{2}$
$\lambda {t_{\dfrac{1}{2}}} = {\log _e}2$
${t_{\dfrac{1}{2}}} = \dfrac{{{{\log }_e}2}}{\lambda }$ , this is the half- life of a radioactive substance.
Now, we will start to find the mean life.
The number of nuclei decay in the time interval t to $t + dt$ is,
$- \dfrac{{dN(t)}}{{dt}} = \left( { - {N_0}\lambda {e^{ - \lambda t}}dt} \right)$
$= {N_0}\lambda {e^{ - \lambda t}}dt$
Each of them has lived for the time t. Thus, the total life of all these nuclei will be t ${N_0}\lambda {e^{ - \lambda t}}dt$. Therefore, to obtain the mean life, we have to integrate this expression over all times from 0 to $\infty$.
So,
Mean life,
$\tau = \dfrac{{\lambda {N_0}\int\limits_0^\infty {t{e^{ - \lambda t}}dt} }}{{{N_0}}}$
$= \lambda \int\limits_0^\infty {t{e^{ - \lambda t}}dt}$
$= \int\limits_0^\infty {\lambda t{e^{ - \lambda t}}dt}$
Since, $\int {x{e^{ - x}}} dx = \dfrac{1}{x}$, which is a mathematical identity.
So, $\tau = \dfrac{1}{\lambda }$ , this is the mean life.

So, the correct option is option (B). ie, $\dfrac{{{{\log }_e}2}}{\lambda }$and $\dfrac{1}{\lambda }$

Note: The mathematical identity used in the solution should be remembered by the students. For solving the actual numerical based on the law of radioactive decay, you must always use the SI units of all the physical quantities.
In case we want to narrow down our options, we can also keep an eye out for the dimension of the quantities. Here in options C and D. The quantities are of dimension $[{T^{ - 1}}]$ and we want it to be $[{T^1}]$. Hence C and D are gone. Next we have A and B. We know that half life is smaller than mean life so half life should be $\dfrac{{{{\log }_e}2}}{\lambda }$ and mean life should be $\dfrac{1}{\lambda }$. Hence option B is correct