Question

The half-life of a radioactive substance is 100 years. The number of years after which the activity will decay to $\dfrac{1}{{10}}th$ ​of its initial value is:
(A) 250 years
(B) 300 years
(C) 333.3 years
(D) 350.3 years

Answer 1

As per the question, first we will assume the activity and initial activity be $R$ and ${R_0}$ respectively, and then we will apply the formula in the terms of of activities and the half-life of radioactive substance, i.e. $\therefore \dfrac{{0.639}}{{{T_{\dfrac{1}{2}}}}} \times t = \ln \dfrac{{{R_0}}}{R}$ , and we will get the required time.

Complete Step By Step Answer:
According to the question, the activity will decay to $\dfrac{1}{{10}}th$ ​of its initial value that means:
$\because R = \dfrac{{{R_0}}}{{10}}$
where, $R$ is the actual or final activity and
${R_0}$ is the initial activity.
And also given that the half-life of a radioactive substance is 100 years:
$\because {T_{\dfrac{1}{2}}} = 100\,years$
where, ${T_{\dfrac{1}{2}}}$ is the half-life of the given substance.
Now, we will use the formula in terms of activities and the half-life of radioactive substances:-
$\therefore \dfrac{{0.639}}{{{T_{\dfrac{1}{2}}}}} \times t = \ln \dfrac{{{R_0}}}{R}$
where, $t$ is the time period.
So, now we will put the given above values:-
$\Rightarrow \dfrac{{0.693}}{{100}} \times t = \ln (10) \\\ \Rightarrow t = \dfrac{{100}}{{0.693}} \times 2.303 \\\ \Rightarrow t = 333.3\,years \\\$
Therefore, 333.3 years after which the activity will decay to $\dfrac{1}{{10}}th$ ​of its initial value.
Hence, the correct option is (C) 333.3 years.

Note:
In radioactivity, the half-life is the time required for one-half of a radioactive sample's atomic nuclei to decay (change spontaneously into other nuclear species by emitting particles and energy), or, equivalently, the time required for a radioactive sample's number of disintegrations per second.