Question

Activity of a radioactive element is ${10^3}\,dps$. Its half-life is $1\,\sec $. After $3\,\sec $, 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 final activity of the radioactive material can be determined by the using the formula of the activity of the radioactive material. The activity of the radioactive material gives the relation between the activity and the time of the decay.

Formula used:
The activity of the radioactive material is given by,
$A = \dfrac{{{A_0}}}{{{2^n}}}$
Where, $A$ is the final activity of the radioactive material, ${A_0}$ is the initial activity of the radioactive material and $n$ is the time taken for the decay.

Complete step by step solution:
Given that,
The initial activity of the radioactive material is, ${A_0} = {10^3}$,
The time taken for the decay, $t = 3\,\sec $.
Now,
The activity of the radioactive material is given by,
$A = \dfrac{{{A_0}}}{{{2^n}}}\,....................\left( 1 \right)$
By substituting the initial activity of the radioactive material and the time taken for the decay in the above equation (1), then the above equation (1) is written as,
\[A = \dfrac{{{{10}^3}}}{{{2^3}}}\]
By taking the cube on the both numerator and the denominator in the above equation, then the above equation is written as,
\[A = \dfrac{{1000}}{8}\]
By dividing the terms in the above equation, then the above equation is written as,
$A = 125\,dps$

Hence, the option (C) is the correct answer.

Note: The final activity of the radioactive material is directly proportional to the initial activity of the radioactive material and inversely proportional to the time of the decay. The initial activity of the radioactive material increases, the final activity of the radioactive material also increases.

Alternate method
The formula for finding the radioactivity is,
$A = {A_0}{e^{ - \lambda T}}$
Where, $A$ is the final activity of the radioactive material, ${A_0}$ is the initial activity of the radioactive material and $T$ is the time taken for the decay.
By using this formula also, the solution can be determined.