Physics - Modern Physics Question with Solution | TestHub

Let the initial amount of both samples be $N_0$.

The amount of radioactive substance remaining after time $t$ is given by $N=N_0{\left(\frac{1}{2}\right)}^{\frac{t}{T_{1/2}}}$.

For sample A, after $t=9$ hours:

$\frac{N_0}{8}=N_0{\left(\frac{1}{2}\right)}^{\frac{9}{T_A}}$

${\left(\frac{1}{2}\right)}^3={\left(\frac{1}{2}\right)}^{\frac{9}{T_A}}$

$\frac{9}{T_A}=3\Rightarrow T_A=3$ hours.

For sample B, after $t=9$ hours:

$\frac{N_0}{128}=N_0{\left(\frac{1}{2}\right)}^{\frac{9}{T_B}}$

${\left(\frac{1}{2}\right)}^7={\left(\frac{1}{2}\right)}^{\frac{9}{T_B}}$

$\frac{9}{T_B}=7\Rightarrow T_B=\frac{9}{7}$ hours.

The ratio of the half-life period of B to that of A is:

$\frac{T_B}{T_A}=\frac{\frac{9}{7}}{3}=\frac{3}{7}$

Answer: $3:7$