Question

Activity of a radioactive element is ${10^{3\,\,dps}}$. Its half-life is $1\,\,s$. After $3\,\,s$ its activity will be ($dps = $decay per second)
(A) $1000\,\,dps$
(B) $250\,\,dps$
(C) $125\,\,dps$
(D) none of these

Answer 1

Hint: The given problem can be solved using the formula of activity of the radioactive material which incorporates the time that the decay takes place, the decay constant and Eyler’s exponential constant with respect to the half time of the radioactive element.

Formulae Used:
The activity of the radioactive element is;
$A = {A_o} \times {e^{ - \lambda T}}$
Where, $A$ denotes the activity of the radioactive element, ${A_o}$ denotes the initial activity of the radioactivity element, $\lambda $ denotes the decay constant of the radioactive element, $T$ denotes the half life time of the radioactive element, $e$ denotes the Eyler’s exponential constant.

Complete step-by-step solution:
The data given in the problem is;
Initial activity of the radioactivity element, ${A_o} = {10^3}\,\,dps$,
Half life time of the radioactive element, $T = 1\,\,s$.
 The activity of the radioactive element is;
$A = {A_o} \times {e^{ - \lambda T}}$
We know that $\lambda = \dfrac{{0.693}}{T}$, since $T = 1\,\,s$, $\lambda = \dfrac{{0.693}}{1}$
Substitute the values of ${A_o}$, $e$ and $\lambda $ in the equation;
After the time $T = 3\,\,s$ the activity of the radioactive element is;
Since Eyler’s exponential constant, $e = 2.7193$
$A = {10^3} \times {2.7193^{ - 0.693 \times 3}}$
On simplifying the above equation, we get;
$\Rightarrow A = {10^3} \times {2.7193^{ - 2.079}}$
$\Rightarrow A = \dfrac{{1000}}{{{{2.7193}^{ - 2.079}}}}$
$\Rightarrow A = \dfrac{{1000}}{8}$
$\Rightarrow A = 125\,\,dps$
Therefore, the activity of the radioactive element after three seconds is $A = 125\,\,dps$.
Hence, the option (C) $A = 125\,\,dps$ is the correct answer.

Note:- Half-lives can range from seconds that is $224$ half-life $ = 55\,\,s$ to millions of years. This has an association for radioactive waste from nuclear power stations which will be demanded to be put down safely for a very long time. For example, potassium of $40$ half-life$ = 1.3 \times {10^9}$ years.