The decay constant of ${}^{14}C$ is

$\lambda =\frac{0.693}{T_{\frac{1}{2}}}=\frac{0.693}{5730 \times 3.16 \times {10}^7}=3.83 \times {10}^{-12}\mathrm{ }{\mathrm{s}}^{-1}$

$\left[\because \mathrm{ }1\mathrm{ }\mathrm{year}=3.17\mathrm{ }\times {10}^7\mathrm{s}\right]$

To find the number of ${}^{14}C$ nuclei in 20 g of burnt wood, we first calculate the number of ${}^{12}C$ nuclei in 20 g of carbon (burnt wood).

Thus $N\left({}^{12}C\right)=\frac{6.02 \times {10}^{23}}{12} \times 20 \simeq {10}^{24}$

Now, assuming that the ratio of ${}^{14}C \mathrm{t}\mathrm{o} {}^{12}C \mathrm{i}\mathrm{s} 1.3 \times {10}^{-12},$ the number of ${}^{14}C$ nuclei in 20g before decay is, $N_0\left({}^{14}C\right)=\left(1.3 \times {10}^{-12}\right) \left({10}^{24}\right)=1.3 \times {10}^{12}$

We thus have for the initial activity of the sample

$R_0=N_0\lambda =\left(1.3 \times {10}^{12}\right) \times \left(3.83 \times {10}^{-12}\right)=4.979\mathrm{ }\mathrm{decay} {\mathrm{s}}^{-1}$

$\approx 5\mathrm{ }\mathrm{decay} {\mathrm{s}}^{-1}$

The age of the sample can now be calculated from the relation,

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

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

Or $t=\frac{1}{\lambda }{\log }_e\left(\frac{R_0}{R}\right)$

It is given that R = $4 decay {\mathrm{s}}^{-1}$ and we have calculated $R_0=5\mathrm{ }\mathrm{decay} {\mathrm{s}}^{-1}$ .

Thus $t=\frac{1}{\lambda }{\log }_e\left(\frac{5}{4}\right)=\frac{{\log }_e\left(1.25\right)}{3.83 \times {10}^{-12}}$

$=\frac{0.223}{3.83 \times {10}^{-12}}=0.58 \times {10}^{11}\mathrm{ }\mathrm{s}$

$=1840 \mathrm{y}\mathrm{e}\mathrm{a}\mathrm{r}$ .