Question

The activity of a radioactive sample is measured as $9750 \,count/min$ at $t = 0$ and $975 \,count/min$ at $t = 5 min$. The decay constant is approx.
A $0.922{\min ^{ - 1}}$
B $0.691{\min ^{ - 1}}$
C $0.461{\min ^{ - 1}}$
D $0.230{\min ^{ - 1}}$

Answer 1

Hint
The radioactive decay constant is defined as the probability of a given unstable nucleus per unit time. It is denoted by $\lambda$, and it can be calculated by using the formula $R = {R_0}{e^{ - \lambda t}}$where, $R$ and $R_0$ are the activities at time $t = 0$ and time $t = t$ respectively. The values of $R$ and $R_0$ are given in the question so by substituting these values and solving the equations, we get the value of decay constant.

Complete step by step answer
we know that the activity of the sample is given as,
$R = {R_0}{e^{ - \lambda t}}$ where, $\lambda$ is the decay constant, $R_0$ is the activity at the time t=0 and R is the activity at time $t$.
$ \Rightarrow \dfrac{R}{{{R_0}}} = {e^{ - \lambda t}}$
Now, taking natural log both sides, we get
$ \Rightarrow \ln \left( {\dfrac{R}{{{R_0}}}} \right) = - \lambda t$
$\Rightarrow \ln \left( {\dfrac{{{R_0}}}{R}} \right) = \lambda t $
$\Rightarrow \lambda = \dfrac{1}{t}\ln \left( {\dfrac{{{R_0}}}{R}} \right)$
Now, converting the natural log to base 10, we get
$ \Rightarrow \lambda = \dfrac{{2.303}}{t}\log \left( {\dfrac{{{R_0}}}{R}} \right)$ … (1)
As it is given that,
Initial activity of the sample is ${R_0} = 9750 \,count/min$
Activity after t = 5min is $R = 975 \,count/min$
Put these values in the equation (1), we get
$\Rightarrow \lambda = \dfrac{{2.303}}{5}\log \left( {\dfrac{{9750}}{{975}}} \right) $
$\Rightarrow \lambda = \dfrac{{2.303}}{5}\log \left( {10} \right)$
$\Rightarrow \lambda = \dfrac{{2.303}}{5} = 0.461{\min ^{ - 1}}$
Hence, (C) option is correct.

Note
The radioactive decay constant is defined as the probability of a given unstable nucleus per unit time. It is denoted by $\lambda$, while calculating the decay constant make sure that the base of log should be $10$ otherwise you will get the wrong answer.