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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: 1og₁₀2 = 0.3010)

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answer is 56.445.

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Detailed Solution

Let t be the time taken by the activity (R) to drop to (1/50) of its present unsafe value (R₀), i.e., $\frac{R}{R_{0}} = \frac{1}{50}$

As $R = λN and R_{0} = {λN}_{0}$

$\frac{N}{N_{0}} = \frac{R}{R_{0}} = \frac{1}{50} or \frac{N_{0}}{N} = 50$

As $\frac{N_{0}}{N} = 2^{t / (T_{1 / 2})} \Rightarrow 50 = 2^{t / 10}$

Thus, $\log 50 = \log 2^{t / 10} or \log 50 = \frac{t}{10} \log 2$

or $t = \frac{10 \times \log 50}{\log 2} = \frac{10 \times 1 .6990}{0 .3010} = 56 .4452 days$

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