Question

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.

Short answer

The activity of the nuclide ${}^{14}\mathrm{C}$is reduced after $3.2\times {10}^{4}y$.

Step-by-step solution

Given data

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 ${}^{14}C,T_{1/2}=5730y$

Understanding the concept of activity of the sample

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, $R=R_0{e}^{-\lambda t}.........\left(2\right)$

Calculation of the time at which the activity of the nuclide is decayed

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 $3.2\times {10}^4\mathrm{y}$.