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

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 $t=\frac{10 \times \log 50}{\log 2}=\frac{10 \times 1.6990}{0.3010}$
$=56.45\, d \approx 56.5$ days