Question

For a certain radioactive process the graph between \(\ln R\) and \(t~\text{(sec)}\) is obtained as shown in the figure. Then the value of half life for the unknown radioactive material is approximately:

InR Nos o ® R = decay rate 10 20 30 40 50 60 time, t(sec)

• A. $4.62\text{sec}$
• B. $2.62\text{sec}$
• C. $9.15\text{sec}$
• D. $6.93\text{sec}$

Answer 1

The correct option is A $4.62\text{sec}$

We know that,

$R=R_0{e}^{-\lambda t}$

$\Rightarrow \ln R=-\lambda t+\ln R_0$

$-\lambda$ is slope of straight line. So, from diagram,

$-\lambda =\frac{6-0}{0-40}=-\frac{3}{20}$

$\Rightarrow \lambda =\frac{3}{20}{\text{s}}^{-1}$

For half life,

$t_{1/2}=\frac{0.693}{\lambda }=\frac{0.693}{3/20}=4.62\text{s}$

Hence, option $\left(\text{D}\right)$ is correct.