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Question

(B) The activity $A$ of a radioactive substance after $n$ half-lives is given by $A = A_0 \left( \frac{1}{2} \right)^n$,where $n = \frac{t}{T_{1/2}}$.

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) The activity $A$ of a radioactive substance after $n$ half-lives is given by $A = A_0 \left( \frac{1}{2} ight)^n$,where $n = \frac{t}{T_{1/2}}$.Given that the activity becomes $1/64$ of the initial activity $A_0$,we have $\frac{A}{A_0} = \frac{1}{64}$.Substituting this into the equation: $\frac{1}{64} = \left( \frac{1}{2} ight)^n$.Since $64 = 2^6$,we can write $\left( \frac{1}{2} ight)^6 = \left( \frac{1}{2} ight)^n$,which gives $n = 6$.Given the half-life $T_{1/2} = 2 	ext{ hours}$,the total time $t$ is $t = n 	imes T_{1/2} = 6 	imes 2 = 12 	ext{ hours}$.

$A$ radioactive substance in a laboratory has a half-life of $2 \text{ hours}$. It is considered safe for laboratory use when its activity reduces to $1/64$ of its initial value. After how many hours will the laboratory be safe?

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