Question

The activity of a sample is ${A_1}$ at a time ${t_1}$ and ${A_2}$ at time ${t_2}$ $({t_2} > {t_1})$. The half-life of the sample is $T$. Then, the number of atoms that have disintegrated in time $({t_1} - {t_2})$ is proportional to
A) $\dfrac{{{A_1}}}{{{A_2}}}$
B) $({A_1} + {A_2})T$
C) $({A_1} - {A_2})T$
D) $\dfrac{{({A_1} - {A_2})}}{T}$

Answer 1

In this question, we can use the formula $A = \lambda N$to find the number of disintegrations for a sample. Then we can find the number of disintegrations in $({t_1} - {t_2})$. We can modify the result of number of disintegrations in $({t_1} - {t_2})$using formula $\lambda = \dfrac{{{{\log }_e}2}}{T}$because we have given half-life of the sample.

Complete step by step solution: -
Let the total number of particles in the sample is $N$ and the half-life of the sample is $T$. Then the decay constant $\lambda$ will be given as-
$\lambda = \dfrac{{{{\log }_e}2}}{T}$ ……………..(i)
We know that the activity $A$ is the number of decays per unit time of a radioactive sample. So, it is given as-
$A = \lambda N$
Or
$N = \dfrac{A}{\lambda }$..................(ii)
Where $N$ is the number of disintegrations in that sample and $\lambda$is the decay constant.
Now, according to the question, the activity of a sample is ${A_1}$ at a time ${t_1}$ and ${A_2}$ at time ${t_2}$.
So, at $t = {t_1}$ , putting $N = {N_1}$ and $A = {A_1}$ in equation (ii) , we get-
${N_1} = \dfrac{{{A_1}}}{\lambda }$....................... (iii)
And at $t = {t_2}$ , putting $N = {N_2}$ and $A = {A_2}$ in equation (ii), we get-
${N_2} = \dfrac{{{A_2}}}{\lambda }$ …………………….(iv)
So, the number of disintegrations is the difference between ${N_1}$ and ${N_2}$. So, subtracting equation (iii) from equation (iv), we get-
${N_1} - {N_2} = \dfrac{{{A_1}}}{\lambda } - \dfrac{{{A_2}}}{\lambda } \\\ \Rightarrow {N_1} - {N_2} = \dfrac{{({A_1} - {A_2})}}{\lambda } \\\$
Here ${N_1}$is subtracting from ${N_2}$because ${t_2}$is greater than ${t_1}$. So, the number of disintegrations ${N_1}$is greater than the number of disintegrations ${N_2}$.The decay constant $\lambda$ is a characteristics for a sample, so it is same for all the $N$.
Now, putting the value of $\lambda$from equation from equation (i), we get-
${N_1} - {N_2} = \dfrac{{({A_1} - {A_2})T}}{{{{\log }_e}2}}$
The half-life $T$ is also a characteristic for a sample. So, it is the same for all $N$.
As ${\log _e}2$ is a constant. So, the equation can be written as-
${N_1} - {N_2} \propto ({A_1} - {A_2})T$
So, the number of atoms$({N_1} - {N_2})$ that have disintegrated in time $({t_1} - {t_2})$ is proportional to$({A_1} - {A_2})T$.

Hence, option C is correct.

Note: - In this question, we can find different numbers of disintegrations for different times. Decay constant $\lambda$and half-life$T$ are the characteristics of a sample and will be equal for that sample at any time of disintegration. ${\log _e}2$is a constant and will be removed from the equation after adding the proportionality symbol.