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Some amount of radioactive substance (half life=10 days) is spread inside a room and consequently the level of radiation becomes 50 times the permissible level for normal occupancy of the room. The room be safe for occupation after

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Detailed Solution

Since the initial activity is 50 times the activity for safe occupancy, therefore , R₀ = 50R where R = $λ$ N since,

$\begin{array}{l} R \propto N \\ \frac{R}{R_{0}} = \frac{N}{N_{0}} = {(\frac{1}{2})}^{n} = {(\frac{1}{2})}^{t / T} or \\ {(\frac{1}{2})}^{t / 10} = \frac{1}{50} \end{array}$

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