Physics - Modern Physics Question with Solution | TestHub

The activity of a radioactive sample at time $t$ is given by the radioactive decay law:

$A=A_0{e}^{-\lambda t}$

Given that the activity decreases to $\frac{1}{3}$ of its original value in $3$ days, we have:

$\frac{A_0}{3}=A_0{e}^{-3\lambda }$

$e^{-3\lambda }=\frac{1}{3}$

We need to find the activity after $9$ days. Substituting $t=9$ into the decay equation:

$A^{\prime }=A_0{e}^{-9\lambda }$

$A^{\prime }=A_0(e^{-3\lambda }{)}^3$

Substituting the value of $e^{-3\lambda }$:

$A^{\prime }=A_0{\left(\frac{1}{3}\right)}^3$

$A^{\prime }=\frac{A_0}{27}$

Thus, the activity will become $\left(\frac{1}{27}\right)$ of the original value.

Answer: $\left(\frac{1}{27}\right)$ of the original value