Chapter 31, Problem 64GP

To determine

The activity Chernobyl radiation per square meter.

Answer to Problem 64GP

  $1.45\times {10}^3\text{ decay/s}\cdot {\text{m}}^2$

Explanation of Solution

Given:

Radiation released $=\text{2}\times {\text{10}}^7\text{Ci}$

  $1\text{ Ci}=3.70\times {10}^{10}\text{ dis/s}$

Surface area of earth $=5.10\times {10}^{14}{\text{ m}}^2$

Formula used:

Activity in square meter on surface of earth $=\frac{\text{Activity}}{\text{Surface area of earth}}$

Calculation:

Converting the unit of radiation released from $\text{Ci}$ to $\text{dis/s}$ ,

  $\begin{array}{l}\text{2 Ci}=\left(3.7\times {10}^{10}\times 2\times {10}^7\right)\text{dis/s}\\ \text{2 Ci}=7.4\times {10}^{17}\text{dis/s}\end{array}$

The activity of Chernobyl per square meter is calculated as,

  $\begin{array}{l}{\text{Activity/m}}^2=\frac{7.4\times {10}^{17}}{5.10\times {10}^{14}{\text{ m}}^2}\\ {\text{Activity/m}}^2=1.45\times {10}^3\text{ decay/s}\cdot {\text{m}}^2\end{array}$

Conclusion:

The activity Chernobyl radiation per square meter is $1.45\times {10}^3\text{ decay/s}\cdot {\text{m}}^2$ .