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

Hint: 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.