The activity of an element becomes frac{1}{64 | vedclass

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Question

(B) The activity $A$ of a radioactive sample at time $t$ is given by the formula: $A = A_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}$,where $A_0$

Vedclass · Accepted Answer

$10$

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(B) The activity $A$ of a radioactive sample at time $t$ is given by the formula: $A = A_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}$,where $A_0$ is the initial activity and $T_{1/2}$ is the half-life.Given that the activity becomes $\frac{1}{64}$ of its original value in $t = 60 \ s$,we have:$\frac{A}{A_0} = \frac{1}{64} = \left( \frac{1}{2} \right)^{\frac{60}{T_{1/2}}}$Since $\frac{1}{64} = \left( \frac{1}{2} \right)^6$,we can equate the exponents:$6 = \frac{60}{T_{1/2}}$$T_{1/2} = \frac{60}{6} = 10 \ s$.Therefore,the half-life period is $10 \ s$.

The activity of an element becomes $\frac{1}{64}$ of its original value in $60 \ s$. Then the half-life period is ............ $s$.

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