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(D) The activity 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 t

Vedclass · Accepted Answer

$62$

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(D) The activity of a radioactive sample at time $t$ is given by the formula $A = A_0 \left( \frac{1}{2} ight)^{\frac{t}{T_{1/2}}}$,where $A_0$ is the initial activity,$A$ is the final activity,$t$ is the time elapsed,and $T_{1/2}$ is the half-life.Given: $A_0 = 700 \; s^{-1}$,$A = 500 \; s^{-1}$,and $t = 30 \; min$.Substituting the values: $500 = 700 \left( \frac{1}{2} ight)^{\frac{30}{T_{1/2}}}$.Dividing both sides by $700$: $\frac{5}{7} = \left( \frac{1}{2} ight)^{\frac{30}{T_{1/2}}}$.Taking the natural logarithm on both sides: $\ln(5/7) = \frac{30}{T_{1/2}} \ln(1/2)$.$\ln(0.714) = \frac{30}{T_{1/2}} (-0.693)$.$-0.337 = \frac{30}{T_{1/2}} (-0.693)$.$T_{1/2} = \frac{30 	imes 0.693}{0.337} \approx 61.69 \; min$.Rounding to the nearest integer,the half-life is approximately $62 \; min$.

The activity of a radioactive sample falls from $700 \; s^{-1}$ to $500 \; s^{-1}$ in $30 \; min$. Its half-life is close to ......... $min$.

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