Question

The activity of a sample of a radioactive material is ${A_1}$ , at time ${t_1}$ and ${A_2}$ at time ${t_2}$, ${({t_2} > {t_1})}$. If its mean life T , then
A. ${{A_1}{t_1} = {A_2}{t_2}}$
B. ${{A_1} - {A_2} = {t_2} - {t_1}}$
C. ${{A_2} = {A_1}{e^{({t_1} - {t_2})/T}}}$
D. ${{A_2} = {A_1}{e^{({t_1}/{t_2})T}}}$

Answer 1

Hint:We will use the law of radioactive disintegration. Then we simplify the equation using the law of radioactive disintegration of radioactive materials.


Formula used:
Law of radioactive disintegration
${N = {N_0}{e^{ - \lambda t}}}$
Where N is the total number of particles in the sample and {N_0}is the initial activity.


Complete answer:
Given data are
Activity at ${t = {t_1}}$ is ${{A_1}}$
Activity at ${t = {t_2}}$ is ${{A_2}}$
Mean life = T
We know that the mean life,
${T = \frac{1}{\lambda },\lambda = coefficient~of~disintegration}$
 ${\Rightarrow \lambda = \frac{1}{\lambda } - 1}$
By the law of radioactive disintegration
Activity , ${{A_0}}$ = initial activity
 $\Rightarrow A = {A_0{e^{ - \lambda {t_1}}}}$ and
 $\Rightarrow {A_2} = {A_0}{e^{ - \lambda {t_2}}}$
  $\Rightarrow \frac{{{A_2}}}{{{A_1}}} = {e^{\lambda {t_1} - \lambda {t_2}}}$
  $\Rightarrow {A_2} = {A_1}{e^{\lambda ({t_1} - {t_2})}} $
Now, we simplify the above equations and we get
${{A_2} = {A_1}{e^{({t_1} - {t_2})/T}}}$

Hence option C is the correct option

Additional information
The law of radioactive disintegration can be applied to find the activity sample of the radioactive materials.



Note:The activity of a sample of radioactive matter is defined by the number of disintegrations taking place at its core at any given moment. The activity also represents the number of radiations emitted.