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Found in: Page 1307

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# 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 ${\mathbf{2}}{\mathbf{.}}{\mathbf{6}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{5}}}{{\mathbf{km}}}^{{\mathbf{2}}}$ was estimated to be ${\mathbf{1}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{16}}}{\mathbf{Bq}}$ . 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. The number of beta-decay electrons that would be intercepted by me if I were to lie on the ground in that area for 1h in 1996 is $7×{10}^{7}$ .
2. The number of beta-decay electrons that would be intercepted by me if I were to lie on the ground in that area for 1h in today is $3.85×{10}^{7}$ .
See the step by step solution

## Step 1: Write the given data

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

## Step 2: Determine the concept of decay

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)

## Step 3: a) Determine the number of electrons intercepted by me in 1996

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

$\frac{\mathrm{rate}\mathrm{throughme}}{\mathrm{total}\mathrm{rate}\mathrm{upward}}=\frac{1{\mathrm{m}}^{2}}{\left(2.6×{10}^{5}{\mathrm{km}}^{2}\right)\left(1000{\mathrm{m}}^{2}/{\mathrm{km}}^{2}\right)}\phantom{\rule{0ex}{0ex}}=3.8×{10}^{-12}$

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:

$\mathrm{rate}\mathrm{through}\mathrm{me}=\frac{1}{2}\left(1×{10}^{16}\frac{1}{\mathrm{s}}\right)\left(3.8×{10}^{-12}\right)\phantom{\rule{0ex}{0ex}}=1.9×{10}^{4}\frac{1}{\mathrm{s}}\phantom{\rule{0ex}{0ex}}\approx 7×{10}^{7}\frac{1}{\mathrm{h}}$

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

## Step 4: b) Determine the number of electrons intercepted by me in today

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:

$\mathrm{R}=\left(7×{10}^{7}{\mathrm{h}}^{-1}\right){\mathrm{e}}^{-\left(\frac{\mathrm{In}2}{30.2}\right)26\mathrm{y}}\phantom{\rule{0ex}{0ex}}=\left(7×{10}^{7}{\mathrm{h}}^{-1}\right){\mathrm{e}}^{-0.59675}\phantom{\rule{0ex}{0ex}}=3.85×{10}^{7}{\mathrm{h}}^{-1}$

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

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