Chapter 13, Problem 13.1P

To determine

The dose rate in the neutron field.

Answer to Problem 13.1P

  $\text{7}\text{.6 Sv/h}$ .

Explanation of Solution

Given:

Mass, $m=500\text{mg}$

Count per minute, $c=500\text{cpm}$

Formula used:

The saturated activity is given by

  $\lambda N=\varphi \sigma \text{n}\left(1-e^{-\lambda t}\right)$

Where,

  $\varphi$ =Neutron flux

  $\lambda$ =Disintegration constant

  $t$ =Time

  $\sigma$ =Cross section

Calculation:

Fast neutron irradiation of ${}^{\text{32}}\text{S}$ produces ${}^{\text{32}}\text{P}$ , whose half-life is ${\text{T}}_{1/2}=14.3\text{days}$

The disintegration constant

  $\lambda =\frac{0.693}{T_{1/2}\left({}^{32}\text{P}\right)}=\frac{0.693}{14.3\text{d}\times 24\text{h/d}}=2\times {10}^{-3}{\text{h}}^{-1}$

This field is counted in a $2\pi$ counter 24h after the end of irradiation. So, induced activity is

  $A=\frac{500\text{counts}}{\min }\times \frac{2\text{dis}}{1\text{count}}\times \frac{1\min }{60\mathrm{s}}=16.67\frac{\text{dis}}{\mathrm{s}}$

So, for 24 hours of decay, activity is

  $\begin{array}{l}A=A_0{e}^{-\lambda t}\\ A_0=\frac{A}{e^{-\lambda t}}=\frac{16.67\text{dps}}{e^{-\left(0.002{\text{h}}^{\text{-1}}\right)\times \left(24\text{h}\right)}}\\ A_0=17.5\text{dps}\end{array}$

The cross section for ${}^{32}\text{S}$ is

  $\sigma =0.4\text{barn}\text{=}\text{0}\text{.4}\times {\text{10}}^{-24}\frac{{\text{cm}}^2}{\text{atom}}$

Number of ${}^{32}\text{S}$ target atom is,

  $\begin{array}{l}n=0.5\text{g}\times \frac{6.023\times {10}^{23}\text{atoms/mole}}{32\text{g/mole}}\\ =9.4\times 10\text{a}\text{toms}\end{array}$

The saturated activity is given by

  $\begin{array}{l}\lambda N=\varphi \sigma \text{n}\left(1-e^{-\lambda t}\right)\\ \varphi =\frac{\lambda N}{\sigma \text{n}\left(1-e^{-\lambda t}\right)}\\ \varphi =\frac{17.5{\text{s}}^{\text{-1}}}{0.4\times {10}^{-24}\frac{{\text{cm}}^{\text{2}}}{\text{atom}}\times 9.4\times {10}^{22}\text{atoms}\left(1-e^{-\left(0.002{\text{h}}^{\text{-1}}\right)\times \left(24\text{h}\right)}\right)}\\ \varphi =1.17\times {10}^6\frac{\text{neutron}}{{\text{cm}}^{\text{2}}\text{.s}}\end{array}$

The flux from other neutrons is given by,

  $\begin{array}{l}\frac{{\varphi }_i}{\varphi \left(\frac{1\text{mSv}}{40\text{h}}\right)}=\frac{H_i\left(\frac{\text{mSv}}{\text{h}}\right)}{\left(\frac{1}{40}\frac{\text{mSv}}{\text{h}}\right)}\\ {\dot{H}}_i=\frac{1}{40}\times \frac{{\varphi }_i}{\varphi \left(\frac{\text{1}\text{mSv}}{40\text{h}}\right)}\end{array}$

Energy	${\varphi }_i\times {10}^6{\text{n/cm}}^{\text{2}}\text{s}$	$\varphi \left(\frac{1\text{mSv}}{40\text{h}}\right)$	$H_i\left(\frac{\text{mSv}}{\text{h}}\right)$
Thermal	4.68	270	433
1000 eV	2.34	280	209
0.01MeV	1.17	290	101
0.1 MeV	1.17	58	504
1 MeV	1.17	10	2925
10 MeV	1.17	8.5	3441

Thus, the dose rate is given by

  $\begin{array}{l}H={\displaystyle \sum H_i}\\ H=\left(433\frac{\text{mSv}}{\text{h}}\right)+\left(209\frac{\text{mSv}}{\text{h}}\right)+\left(101\frac{\text{mSv}}{\text{h}}\right)+\left(504\frac{\text{mSv}}{\text{h}}\right)+\left(2925\frac{\text{mSv}}{\text{h}}\right)+\left(3441\frac{\text{mSv}}{\text{h}}\right)\\ H=7613\frac{\text{mSv}}{\text{h}}\\ H=7.6\frac{\text{Sv}}{\text{h}}\end{array}$

Conclusion:

The dose rate in the neutron field is $\text{7}\text{.6 Sv/h}\text{.}$