Question

In a radioactive material the activity at time $ {t_1} $ is $ {R_1} $ and at a later time $ {t_2} $ , it is $ {R_2} $ . If the decay constant of the material is $ \lambda $ , then:
(A) $ {R_1} = {R_2}{e^{ - \lambda \left( {{t_1} - {t_2}} \right)}} $
(B) $ {R_1} = {R_2}{e^{\lambda \left( {{t_1} - {t_2}} \right)}} $
(C) $ {R_1} = {R_2} = \left( {{t_2} - {t_1}} \right) $
(D) $ {R_1} = {R_2} $

Answer 1

Hint
 The radioactivity of a radioactivity is proportional to the number of nuclei in the sample. If we are measuring the activity we are measuring the number of nuclei that are disintegrating. With time the number of nuclei present in the sample decreases.

Complete step by step answer
The unit of activity is Becquerel (Bq). One Becquerel is the decay rate of one disintegration per one second.
The decay rate or rate of disintegration is the number of decays per second. The expression for the rate of disintegration is given as,
 $ R = - \dfrac{{dN}}{{dt}} $
Where, $ N $ is the number of nuclei in the sample.
And we have the rate of number of nuclei per second is proportional to the number of nuclei. Therefore,
 $ - \dfrac{{dN}}{{dt}} = \lambda N $
Where, $ \lambda $ is the decay constant.
Comparing the both equations, we get
 $ R = \lambda N $
We have $ N = {N_0}{e^{ - \lambda t}} $
Where, $ N $ is the number of nuclei at time $ t $ and $ {N_0} $ is the number of nuclei at $ t = 0 $ .
Substituting this in the equation for activity gives,
 $ R = \lambda {N_0}{e^{ - \lambda t}} $
 $ = {R_0}{e^{ - \lambda t}} $
Where, $ \lambda {N_0} = {R_0} $ . And $ {R_0} $ is the activity at time $ t = 0 $ .
Therefore, $ R = {R_0}{e^{ - \lambda t}} $
For time $ {t_1} $ , $ {R_1} = {R_0}{e^{ - \lambda {t_1}}}...........\left( 1 \right) $
And for time $ {t_2} $ , $ {R_2} = {R_0}{e^{ - \lambda {t_2}}}...............\left( 2 \right) $
Dividing equation $ \left( 1 \right) $ and $ \left( 2 \right) $ , we get
 $ \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{R_0}{e^{ - \lambda {t_1}}}}}{{{R_0}{e^{ - \lambda {t_2}}}}}$
 $ \dfrac{{{R_1}}}{{{R_2}}} = {e^{ - \lambda \left( {{t_1} - {t_2}} \right)}} $
  ${R_1} = {R_2}{e^{ - \lambda \left( {{t_1} - {t_2}} \right)}} $
The answer is option A.

Note
If the radioactivity is greater means shorter will be the half -life of the material. The activity can be varied according to the time. That is the radiation emitted will vary in time.