How many years are needed to reduce the activity of${}^{\mathbf{14}}\mathbf{C}$to 0.020of its original activity? The half-life of${}^{\mathbf{14}}\mathbf{C}$is 5730 y.

The activity of the remaining undecayed nuclei, $R=0.02R_0$ where, $R_0$is the initial activity of the nuclide ${}^{14}\mathrm{C}$.

Half-life of nuclide ${}^{14}C,T_{1/2}=5730y$

The activity of a sample is the average number of disintegrations per second. It is nothing but the rate of decay of the nuclei present in the sample. Thus, it is given by initial activity changing with the exponential function of disintegration constant and time of decay.

Formula:

The disintegration constant,$\lambda =\frac{In2}{T_{1/2}}...........\left(1\right)$

where,$T_{1/2}$is the half-life of the substance.

The activity of the undecayed nuclei of the sample, $R=R_0{e}^{-\lambda t}.........\left(2\right)$

After substituting value of equation (1) in equation (2), the time at which activity of the sample will be reduced by 0.02 can be given as follows:

$$\begin{gathered} t=-\frac{1}{\mathrm{\lambda }}\mathrm{In}\left(\frac{\mathrm{R}}{{\mathrm{R}}_0}\right) \\ =-\left(\frac{{\mathrm{T}}_{1/2}}{\mathrm{In}2}\right)\mathrm{In}\left(\frac{\mathrm{R}}{{\mathrm{R}}_0}\right) \\ =-\left(\frac{5730\mathrm{y}}{\mathrm{In}2}\right)\mathrm{In}\left(\frac{0.02{\mathrm{R}}_0}{{\mathrm{R}}_0}\right) \\ =3.2\times {10}^4\mathrm{y} \end{gathered}$$

Hence, the time of decay is $3.2\times {10}^4\mathrm{y}$.