The activity of a radioactive sample falls from $600{\text{s}}^{-1}$ to $500{\text{s}}^{-1}$ in $40\text{minutes}$. Which of the following statement(s) is(are) true? $(\ln 1.2=0.1823)$

The correct option is D $\text{The half life period is}152\text{min}$.
We know that,

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

Here $A_0=600{\text{s}}^{-1}$, $A=500{\text{s}}^{-1}$

$\Rightarrow 500=\left(600\right)e^{-\lambda t}$

$\Rightarrow e^{-\lambda t}=\frac{5}{6}$

$\Rightarrow \lambda t=\ln \left(6/5\right)$

$\Rightarrow \lambda =\frac{\ln \left(6/5\right)}{t}=\frac{0.1823}{40\text{min}}$

$\Rightarrow \lambda =0.00455{\text{min}}^{-1}$

The half-life period is given by,

$t_{1/2}=\frac{\ln 2}{\lambda }$

$\Rightarrow t_{1/2}=\frac{\ln 2}{0.00455}$

$\Rightarrow t_{1/2}\approx 152\text{min}$

Hence, option $\left(B\right)$ and $\left(D\right)$ are correct.