Question

The count rate from a radioactive sample falls from 4.0 × 10⁶ per second to 1.0 × 10⁶ per second in 20 hours. What will be the count rate 100 hours after the beginning?

Answer 1

Given:
Initial count rate of radioactive sample, A₀ = 4 × 10⁶ disintegration/sec
Count rate of radioactive sample after 20 hours, A = 1 × 10⁶ disintegration/sec
Time, t = 20 hours
Activity of radioactive sample $\left(A\right)$ is given by
$$\begin{gathered} A=\frac{A_0}{2^{{\displaystyle \frac{t}{{\displaystyle \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}}} \\ \mathrm{Here},T_{1/2}=\mathrm{Half}-\mathrm{life}\mathrm{period} \\ \mathrm{On}\mathrm{substituting}\mathrm{the}\mathrm{values}\mathrm{of}{A}_0\mathrm{and}A,\mathrm{we}\mathrm{have} \\ {2}^{\frac{t}{{\displaystyle \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}=2^2 \\ \Rightarrow \frac{t}{{\displaystyle \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=2 \\ \Rightarrow \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=t/2=20\mathrm{h}/2=10\mathrm{h} \end{gathered}$$

100 hours after the beginning,
$$\begin{gathered} \mathrm{Count}\mathrm{rate},A"=\frac{A_0}{2^{{\displaystyle \frac{t}{{\displaystyle \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}}} \\ \Rightarrow A"=\frac{4\times {10}^6}{2^{100/10}} \\ =0.390625\times {10}^4 \\ =3.9\times {10}^3\mathrm{disintegrations}/\sec \end{gathered}$$