For a certain radioactive process, the graph between InR and t (sec) is obtained as shown in the figure. Then, the value of half-life for the unknown radioactive material is approximately

According to law of radioactive decay,

$\mathrm{R}={\mathrm{R}}_0{\mathrm{e}}^{-\mathrm{\lambda t}}$                 ...(i)

where, R is remaining amount, R₀ is initial amount and $\lambda$ is decay constant.

Taking natural log on both sides of Eq. (i), we get

$$\begin{gathered} \ln \mathrm{R}=\ln {\mathrm{R}}_0+\ln {\mathrm{e}}^{-\mathrm{\lambda t}} \\ \Rightarrow \ln \mathrm{R}=\ln {\mathrm{R}}_0-\mathrm{\lambda t} ...\left(\mathrm{ii}\right) \end{gathered}$$

So, for a given graph between ln R and t, $-\mathrm{\lambda }$ is slope.

According to above graph given in question at t = 0, ln R = 6

$\therefore$ Eq. (ii) becomes

ln R₀ = 6                    ...(iii)

and at t = 40 s, ln R = 0

$\therefore$ Eq. (ii) becomes,

$\mathrm{\lambda }=\frac{\ln {\mathrm{R}}_0}{\mathrm{t}}=\frac{\ln {\mathrm{R}}_0}{40}$                 ...(iv)

From Eqs. (iii) and (iv), we get

$\mathrm{\lambda }=\frac{6}{40}$                        ...(v)

$\because$ We know that,

Half-life,          ${\mathrm{t}}_{1/2}=\frac{0.693}{\mathrm{\lambda }}$            ...(vi)

$\therefore$ From Eq. (v) and (vi), we get

$\begin{array}{r}{\mathrm{t}}_{1/2}=\frac{0.693\times 40}{6}\\ \Rightarrow {\mathrm{t}}_{1/2}=4.62\mathrm{s}\end{array}$