Question

If 1 $gm$ of a radioactive substance takes 0.286 $s$ to loose 0.25 $gm$, then half life of sample will be [take $ln\left(3\right)$=1.1 and $ln\left(2\right)$=0.693]

• A. $ln\left(2\right)$ $s$
• B. $2ln\left(3\right)$ $s$
• C. $\frac{5}{ln\left(2\right)}$ $s$
• D. 5 $s$

Answer 1

The correct option is A $ln\left(2\right)$ $s$

By $N=N_0{e}^{-\lambda t}\Rightarrow 0.75=1e^{-\lambda t_0}$
$\frac{75}{100}=e^{-\lambda t_0}\Rightarrow e^{\lambda t_0}=\frac{4}{3}$
Let $T$ be the half life of the sample.
$\lambda t_0=\ln \left(4\right)-\ln \left(3\right)\Rightarrow \frac{\ln \left(2\right)}{T}{t}_0=\ln \left(4\right)-\ln \left(3\right)$
$T=\frac{\ln \left(2\right)t_0}{2\ln \left(2\right)-\ln \left(3\right)}=\frac{0.286\ln \left(2\right)}{2\times 0.693-1.1}=\ln \left(2\right)$