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(B) Given,half-life $T_{1/2} = 2 \, h$.Let the initial intensity be $I_0$ and the safe intensity be $I$.We are given that $I_0 = 64I$,so the ratio $\f

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$12$

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(B) Given,half-life $T_{1/2} = 2 \, h$.Let the initial intensity be $I_0$ and the safe intensity be $I$.We are given that $I_0 = 64I$,so the ratio $\frac{I}{I_0} = \frac{1}{64}$.The intensity of radiation follows the law of radioactive decay: $I = I_0 \left( \frac{1}{2} ight)^n$,where $n$ is the number of half-lives.Substituting the values: $\frac{1}{64} = \left( \frac{1}{2} ight)^n$.Since $64 = 2^6$,we have $\left( \frac{1}{2} ight)^6 = \left( \frac{1}{2} ight)^n$,which gives $n = 6$.The total time $t$ is given by $t = n 	imes T_{1/2}$.Therefore,$t = 6 	imes 2 \, h = 12 \, h$.

$A$ freshly-prepared radioactive source of half-life $2 \, h$ 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 this source is ....... $h$.

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