A radioactive sample has activity of 10,000 disintegrations per second
(dps) after 20 hours. After next 10 hours its activity reduces to 5,000 dps.
Find out its half life and initial activity.

Dear Student,

Please find below the solution to the asked query:

The activity of radioactive material is given by 
$A=A_0{e}^{-\lambda t}$
So we get
$$\begin{gathered} 10000=A_0{e}^{-\lambda \left(20\times 3600\right)} \\ and \\ 5000=A_0{e}^{-\lambda \left(30\times 3600\right)} \end{gathered}$$
dividing above equation we get
$$\begin{gathered} 2=e^{-\lambda \times 36000} \\ \Rightarrow \lambda =\frac{\ln 2}{36000}=1.92\times {10}^{-5} \end{gathered}$$
So half life is 
$T_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{\ln 2}{1.93\times {10}^{-5}}=36000 \text{s}=10 \text{h}$


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