Question

Small amount of a radioactive substance (half-life $=10$ days) is spread inside a room and consequently the level of radiation becomes $50$ times the level for normal occupancy of the room. After how many days will the room be safe for occupation? (Given: $\log 2=0.3010$ )

Answer 1

Answer: 56.5

Let $t$ be the time taken by the activity $(R)$ to drop to $(1/50)$
of its present unsafe value $\left(R_0\right)$,
i.e., $\frac{R}{R_0}=\frac{1}{50}$
As $R=\lambda N$ and $R_0=\lambda N_0$
$\frac{N}{N_0}=\frac{R}{R_0}=\frac{1}{50}$
or $\frac{N_0}{N}=50$
As $\frac{N_0}{N}=2^{t/T_{1/2}};50=2^{t/10}$
Thus, $\log 50=\log 2^{t/10}$
or $\log 50=\frac{t}{10}\log 2$
Or $t=\frac{10\times \log 50}{\log 2}=\frac{10\times 1.6990}{0.3010}$
$=56.45d\approx 56.5$ days