Question

After $300$ days, the activity of a radioactive sample is $5000$ dps (disintegrations per second). The activity becomes ${2500\,dps}$ after another $150$ days. The initial activity of the sample in dps is
(a) $20,000$
(b) $10,000$
(c) $7,000$
(d) $25,000$

Answer 1

In this solution, we are going to use the relation between initial activity and activity after a certain given time. Our aim is to first find out half life ${T_{{1 {\left/{\vphantom {1 2}} \right.} 2}}}$ and then with the help of it, we will find the initial activity of the sample. We would also need to find the number of half lives completed in a time period of $300\;days$.

Complete step by step answer:
Given:
Initial time ${t_1} = 300$ days,
Initial activity ${R_1} = 5000$ dps (disintegrations per second)
Final time ${t_2} = {t_1} + 150$ days,
Final activity ${R_2} = 2500$ dps (disintegrations per second)
Now, we can clearly observe from given data above that, as the time reaches from ${t_1}$ to ${t_2}$ the activity of the sample decreases from $5000$ dps to $2500$ dps, that is, in a time period of $150$ days, the activity of sample reduces to half.
And we know that, the time period in which the activity of a sample decreases to half, is called half life${T_{{1 {\left/{\vphantom {1 2}} \right.} 2}}}$.
Therefore $({t_2} - {t_1}) = 150$days will be the half life ${T_{{1 {\left/{\vphantom {1 2}} \right.} 2}}}$ of the sample.
Mathematically, we can write it as, ${T_{{1 {\left/{\vphantom {1 2}} \right.} 2}}} = 150$days
Number of half lives $n$ completed up to time ${t_1}$ is given by:

$$n = \dfrac{{{t_1}}}{{{T_{{1 {\left/{\vphantom {1 2}} \right.} 2}}}}} \\\ \Rightarrow n = \dfrac{{300\;days}}{{150\;days}} \\\ \Rightarrow n = 2 \\\$$

Let us assume the initial activity of the sample to be${R_0}$.
Then, the formula to calculate activity ${R_0}$ is given as:
$\dfrac{R}{{{R_0}}} = {\left( {\dfrac{1}{2}} \right)^n}\;\;\;........(1)$
Substituting $R = 5000\;dps$ and$n = 2$, in equation (1), we get,
$\dfrac{{5000\;dps}}{{{R_0}}} = {\left( {\dfrac{1}{2}} \right)^2} \\\ \Rightarrow\dfrac{{5000\;dps}}{{{R_0}}} = \left( {\dfrac{1}{4}} \right) \\\ \Rightarrow{R_0} = 5000 \times 4 \\\ \therefore{R_0} = 20,000\;dps$

Therefore, initial activity of the sample is ${20,000\;dps}$. So the option (a) is the correct answer.

Note: While putting the values of $t$ and ${T_{{1 {\left/{\vphantom {1 2}} \right. } 2}}}$ to find the value of$n$, make sure to put the same value of time in $t$ corresponding to the activity value which you are putting in $R$.