Because of the 1986 explosion and fire in a reactor at the Chernobyl nuclear power plant in northern Ukraine, part of Ukraine is contaminated with ${}^{\mathbf{137}}\mathbf{Cs}$,which undergoes beta-minus decay with a half-life of 30.2y. In 1996, the total activity of this contamination over an area of $\mathbf{2}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}{\mathbf{km}}^{\mathbf{2}}$was estimated to be $\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{16}}\mathbf{Bq}$. Assume that the${}^{\mathbf{137}}\mathbf{Cs}$is uniformly spread over that area and that the beta-decay electrons travel either directly upward or directly downward. How many beta-decay electrons would you intercept were you to lie on the ground in that area for(a) in 1996 and (b) today? (You need to estimate your cross-sectional area that intercepts those electrons.)

1. $\mathrm{R}=1\times {10}^{16}\mathrm{Bq}$Half-life of ${}^{137}\mathrm{Cs}$, $T_{1/2}=30.2y$
2. Total activity of the contamination, $\mathrm{R}=1\times {10}^{16}\mathrm{Bq}$
3. Area of contamination, $\mathrm{A}=2.6\times {10}^5{\mathrm{km}}^2$
4. Time of contamination, t = 1 h

The abundance of decay defines as the decay acting in the portion. That is the fraction of the portion at which the decay is happening respectively to the whole ground area. Similarly, using this concept, we can get the rate of contamination passing through my body. Now, using this data, and the time of decay, we can get further contamination after the given time within my body.

The activity of the undecayed nuclei with the given time is as follows:

$R=R_0{e}^{-\left(\frac{In2}{T_{1/2}}\right)t}$ …… (i)

Assuming a “target” area of one square meter, we establish a ratio:

$$\begin{gathered} \frac{\mathrm{rate}\mathrm{throughme}}{\mathrm{total}\mathrm{rate}\mathrm{upward}}=\frac{1{\mathrm{m}}^2}{\left(2.6\times {10}^5{\mathrm{km}}^2\right)\left(1000{\mathrm{m}}^2/{\mathrm{km}}^2\right)} \\ =3.8\times {10}^{-12} \end{gathered}$$

The SI unit Becquerel is equivalent to a disintegration per second. With half the beta-decay electrons moving upward, the total rate through me can be given as:

$$\begin{gathered} \mathrm{rate}\mathrm{through}\mathrm{me}=\frac{1}{2}\left(1\times {10}^{16}\frac{1}{\mathrm{s}}\right)\left(3.8\times {10}^{-12}\right) \\ =1.9\times {10}^4\frac{1}{\mathrm{s}} \\ \approx 7\times {10}^7\frac{1}{\mathrm{h}} \end{gathered}$$

Hence, in one hour$7\times {10}^7$ electrons would be intercepted by me.

For the current year 2022, thus the time of decay becomes

t = (2022 - 1996)

= 26y

Now, using the given data in equation (i), determine the rate of decay in the present given time as follows:

$$\begin{gathered} \mathrm{R}=\left(7\times {10}^7{\mathrm{h}}^{-1}\right){\mathrm{e}}^{-\left(\frac{\mathrm{In}2}{30.2}\right)26\mathrm{y}} \\ =\left(7\times {10}^7{\mathrm{h}}^{-1}\right){\mathrm{e}}^{-0.59675} \\ =3.85\times {10}^7{\mathrm{h}}^{-1} \end{gathered}$$

Hence, the number of electrons intercepted by me in one hour is$3.85\times {10}^7$ .