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(B) The intensity of radiation $I$ follows the law $I = I_0 \left( \frac{1}{2} \right)^n$, where $n$ is the number of half-lives.Given that the initia

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$12$

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(B) The intensity of radiation $I$ follows the law $I = I_0 \left( \frac{1}{2} ight)^n$, where $n$ is the number of half-lives.Given that the initial intensity $I_0 = 64 	imes I_{	ext{safe}}$, we want to find the time when $I = I_{	ext{safe}}$.Substituting the values: $I_{	ext{safe}} = 64 	imes I_{	ext{safe}} 	imes \left( \frac{1}{2} ight)^n$.This simplifies to $\left( \frac{1}{2} ight)^n = \frac{1}{64}$.Since $64 = 2^6$, we have $\left( \frac{1}{2} ight)^n = \left( \frac{1}{2} ight)^6$, which implies $n = 6$.The total time $t = n 	imes T_{1/2}$, where $T_{1/2} = 2\, hours$.Therefore, $t = 6 	imes 2\, hours = 12\, hours$.

$A$ freshly prepared radioactive source of half-life $2\, hours$ emits radiation of intensity which is $64\, \times$ the permissible safe level. The minimum time after which it would be possible to work safely with the source is .......... $hours$.

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