Chapter 31, Problem 66GP

To determine

To calculate: The effective dose received during lab.

Answer to Problem 66GP

  $5.48\times {10}^{-5}\text{rem}$

Explanation of Solution

Given:

Activity of the source, $R=1\mu \text{Ci}$

  $\begin{array}{l}R=1\times {10}^{-6}\times 3.7\times {10}^{10}\\ R=3.7\times {10}^4\text{deays/s}\end{array}$

Formula used:

  $N_o=\frac{R{T}_{1/2}}{0.693}$

Where, $N_o$ = initial amount

  $T_{1/2}$ = half life

Calculation:

  $N_o=\frac{R{T}_{1/2}}{0.693}$

Plugging the values

  $\begin{array}{l}{N}_o=\frac{\left(3.7\times {10}^4\right)\left(30\right)\left(3.153\times {10}^7\right)}{0.693}\\ N_o=5.05\times {10}^{13}\text{atoms}\end{array}$

Number of decays in 2 hr can be calculated as:

  $\begin{array}{l}n=N_o-N\\ n=N_o-N_o{e}^{-\partial t}\\ n=5.05\times {10}^{13}\left(1-e^{-\left(\frac{0.693}{30}\right)\left(2\right)\left(\frac{1}{8760}\right)}\right)\\ n=2.66\times {10}^8\text{decay}\end{array}$

Total energy per decay is:

  $\begin{array}{l}E=190+660\\ E=850\text{keV}\end{array}$

Total energy absorbed per person is:

  $\begin{array}{l}E=\left(2.66\times {10}^8\right)\left(850\right)\left(1.6\times {10}^{-16}\right)\\ E=3.62\times {10}^{-5}\text{J}\end{array}$

Dose in rem is:

  $\begin{array}{l}=dose\left(\text{in rad}\right)\times QF\\ =\frac{3.62\times {10}^{-5}}{66}\times 100\\ =5.48\times {10}^{-5}\text{rem}\end{array}$

Conclusion:

  $5.48\times {10}^{-5}\text{rem}$ is the effective dose received during lab.