
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
163.8k+ views
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.
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.
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