Question

A 300-day radioactive sample shows an activity of 5000 dps. 300 days later, its activity reduces to 2500 dps. The initial activity of the sample is
(1) 2500 dps
(2) 15000 dps
(3) 10000 dps
(4) 12500 dps

Answer 1

Hint: The problems based on the radioactive decay can be solved by knowing the basic concepts of the same. Determining the first order decay equation can help solve this problem.
Radioactive decay is the breakdown of molecules into ‘n’ number of molecules which would result in radiations.

Complete answer:
Let us first learn about the radioactive theory;
Radioactive decay is the spontaneous breakdown of the radioactive atomic nucleus emitting radiations as a result.
The nuclide which undergoes radioactive decay is known as parent nuclide which breaks down to produce daughter nuclides.
The radioactive formula can be expressed as,
     \[R={{R}_{0}}{{e}^{-\lambda T}}\]
where,
${{R}_{0}}$ = initial quantity of the substance
R = the quantity which is remaining from the decay
T = half life of the decaying quantity
e = Euler’s number which is equal to 2.17828
$\lambda $ = decay rate
‘-ve’ sign implies the decay process.
Solving the above equation, we can say that,
     \[\begin{align}
  & \ln \dfrac{R}{{{R}_{0}}}=-\lambda T \\
 & \ln \dfrac{{{R}_{0}}}{R}=\lambda T \\
\end{align}\]
Thus,
The given illustration can now be solved;
Given data-
${{T}_{1}}$ = 300 days
${{R}_{1}}$ = 5000 dps
${{T}_{2}}$ = ${{T}_{1}}$ + 300 = 600 days
${{R}_{2}}$ = 2500 dps
Let, initial activity be ${{R}_{0}}$ dps.
Now, solving as mentioned above,
$\ln \dfrac{{{R}_{0}}}{5000}=\lambda \times 300$
$\ln \dfrac{{{R}_{0}}}{2500}=\lambda \times 600$
Solving above two equations, we get,
$\begin{align}
  & \dfrac{300}{600}=\dfrac{\ln {}^{{{R}_{0}}}/{}_{5000}}{\ln {}^{{{R}_{0}}}/{}_{2500}} \\
 & \dfrac{1}{2}=\dfrac{\ln {}^{{{R}_{0}}}/{}_{5000}}{\ln {}^{{{R}_{0}}}/{}_{2500}} \\
\end{align}$
Thus,
$\begin{align}
  & \dfrac{1}{2}\ln \dfrac{{{R}_{0}}}{2500}=\ln \dfrac{{{R}_{0}}}{5000} \\
 & \ln {{\dfrac{{{R}_{0}}}{2500}}^{{}^{1}/{}_{2}}}=\ln \dfrac{{{R}_{0}}}{5000} \\
 & {{\dfrac{{{R}_{0}}}{2500}}^{{}^{1}/{}_{2}}}=\dfrac{{{R}_{0}}}{5000} \\
\end{align}$
Now, squaring both the sides we get,
$\begin{align}
  & \dfrac{{{R}_{0}}}{2500}={{\dfrac{{{R}_{0}}}{5000}}^{2}} \\
 & \therefore {{R}_{0}}=\dfrac{{{5000}^{2}}}{2500}=10000 \\
\end{align}$
Therefore, the initial activity is 10000 dps.

Hence, option (3) is correct.

Note:
Do note that the given problem is purely based on the radioactive decay and its equation. So, solve and analyse the problem properly.
No remarkable units are mentioned here, then too take specific care if any conversion unit is mentioned.