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(B) The intensity of radiation $I$ decreases according to the law of radioactive decay: $I = I_0 \left( \frac{1}{2} \right)^n$, where $n$ is the numbe

Vedclass · Accepted Answer

$12$

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(B) The intensity of radiation $I$ decreases according to the law of radioactive decay: $I = I_0 \left( \frac{1}{2} ight)^n$, where $n$ is the number of half-lives.Given that the initial intensity $I_0 = 64 I_{	ext{safe}}$, we need the intensity to reach $I_{	ext{safe}}$.Thus, $\frac{I_{	ext{safe}}}{64 I_{	ext{safe}}} = \left( \frac{1}{2} ight)^n$.$\frac{1}{64} = \left( \frac{1}{2} ight)^n$.Since $64 = 2^6$, we have $\left( \frac{1}{2} ight)^6 = \left( \frac{1}{2} ight)^n$, which implies $n = 6$.The total time $t$ is given by $t = n 	imes T_{1/2}$, where $T_{1/2} = 2 	ext{ hours}$.$t = 6 	imes 2 = 12 	ext{ hours}$.

From a newly formed radioactive substance (half-life $2$ hours), the intensity of radiation is $64$ times the permissible safe level. The minimum time after which work can be done safely from this source is .......... $hours$.

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