Physics - Modern Physics Question with Solution | TestHub

Let the initial activity be $A_0$ and the half-life be $T_{1/2}$.

The activity after time $t$ is given by $A=A_0{\left(\frac{1}{2}\right)}^{\frac{t}{T_{1/2}}}$.

Given that $A=\frac{A_0}{4}$ at $t=4$ days:

$\frac{A_0}{4}=A_0{\left(\frac{1}{2}\right)}^{\frac{4}{T_{1/2}}}$

${\left(\frac{1}{2}\right)}^2={\left(\frac{1}{2}\right)}^{\frac{4}{T_{1/2}}}$

$\frac{4}{T_{1/2}}=2\Rightarrow T_{1/2}=2$ days.

After $t=16$ days, the activity $A^{\prime }$ is:

$A^{\prime }=A_0{\left(\frac{1}{2}\right)}^{\frac{16}{2}}=A_0{\left(\frac{1}{2}\right)}^8=\frac{A_0}{256}$

Thus, the activity becomes $\frac{1}{256}$ times the original value, so $x=\frac{1}{256}$.

Answer: $\frac{1}{256}$