Chapter 10, Problem 10.1P

(a)

To determine

The minimum thickness of the paraffin shield.

Answer to Problem 10.1P

The minimum thickness of the paraffin shield is 44 cm.

Explanation of Solution

Given:

Source emits, $S={10}^7\frac{\text{neutron}}{\text{s}}$

The fast neutron flux, $\varphi =10\frac{\text{neutrons}}{{\text{cm}}^{\text{2}}\text{.}\text{s}}$

Attenuation coefficient, $\mu =0.126{\text{cm}}^{\text{-1}}$

Build up factor for hydrogenous shields, B = 5

Formula used:

  $\varphi =\frac{BS}{4\pi t^2}{e}^{-\mu t}$

Where,

t = thickness of paraffin shield

Calculation:

The shield's surface

  $\varphi =\frac{BS}{4\pi t^2}{e}^{-\mu t}$

$10\frac{\text{n}}{{\text{cm}}^2\cdot \text{sec}}=\frac{5\times {10}^7\frac{\text{neutron}}{\text{sec}}}{4\pi t^2}{e}^{-0.126t}$

Using the method of iteration,

First, we assume a small value of t, and then calculating the resulting value $\varphi$ , and then we repeated the process until the value for t is found to be nearest to the desired value of $\varphi$ .

Conclusion:

Att = 44 cm results in $\varphi =8\frac{\text{neutron}}{{\text{cm}}^{\text{2}}\text{.sec}}$ which is close to the desired maximum of $\varphi =10\frac{\text{neutron}}{{\text{cm}}^{\text{2}}\text{.sec}}$ .

(b)

To determine

Thermal neutron leakage atthe surface of the field.

Answer to Problem 10.1P

The thermal neutron leakage is $0.08\frac{\text{n}}{{\text{cm}}^2\text{sec}}$

Explanation of Solution

Given:

Point source emit, S = 10⁷ n/sec

Radius, R = 44 cm

Diffusion coefficient, D = 0.381 cm

thermal diffusion length, L = 3cm

Formula used:

The thermal neutron leakage

  $\varphi =\frac{S{e}^{-R/L}}{4\pi RD}$

Calculation:

The thermal neutron leakage:

  $\varphi =\frac{S{e}^{-R/L}}{4\pi RD}$

${\varphi }_{th}=\frac{{10}^7\frac{\text{n}}{\text{sec}}}{4\times \pi \times 44\text{cm}\times 0.381\text{cm}}{e}^{-\frac{44\text{cm}}{3\text{cm}}}$

  ${\varphi }_{th}=0.08\frac{\text{n}}{{\text{cm}}^2\text{sec}}$

Conclusion:

The thermal neutron leakage is $0.08\frac{\text{n}}{{\text{cm}}^2\text{sec}}$

(c)

To determine

The gamma ray dose rate, due to the hydrogen capture gammas, atthe surface of the shield.

Answer to Problem 10.1P

  $1.4\frac{\text{mrads}}{\text{hr}}$

Explanation of Solution

Given:

Point source emit, S = 10⁷ n/sec

Radius, R = 44 cm

The flux of fast neutron, ${\varphi }_{\text{fast}}=8\frac{\text{n}}{{\text{cm}}^2\text{sec}}$

The flux of thermal neutron, ${\varphi }_{\text{thermal}}=0.08\frac{\text{n}}{{\text{cm}}^2\text{sec}}$

Formula used:

The concentration of gamma emitter

$C=\frac{S-4\pi R^2\left({\varphi }_{\text{fast}}+{\varphi }_{\text{temal}}\right)}{\frac{4}{3}\pi R^3}$

  ${\varphi }_{\text{fast}}$ = flux of fast neutron

  ${\varphi }_{\text{thermal}}$ = flux of thermal neutron

The source strength

  $\Gamma =3.69\times {10}^{-9}{\displaystyle \sum f_i}{E}_i\frac{(\text{C}/\text{kg}){\text{m}}^2}{\text{MBq}\cdot \text{h}}\times 34\frac{\text{Gy}}{\text{C}/\text{kg}}$

Where,

f_i = fraction of the transformations that yield a photon of the ith energy, and

E_i = energy of the i^th photon, MeV

  ${\mu }_1={\mu }_{\text{m}}\times \rho$

Where,

  $\begin{array}{l}{\mu }_l=\text{linear}\text{energy}\text{absoption}\text{coefficient}\text{of}\text{paraffin}\text{.}\\ {\mu }_m=\text{linear}\text{energy}\text{absoption}\text{coefficient}\text{of}\text{water}\text{.}\end{array}$

The dose rate :

  $\dot{D}=\frac{1}{2}C\Gamma \frac{4\pi }{\mu }\left(1-e^{-\mu R}\right)$

Calculation:

Concentration of gamma emitter

$C=\frac{S-4\pi r^2\left({\varphi }_{\text{fast}}+{\varphi }_{\text{temal}}\right)}{\frac{4}{3}\pi r^3}$

  $C=\frac{{10}^7\frac{\text{n}}{\text{s}}-4\pi {(44\text{cm})}^2(8+0.08)\frac{\text{n}}{{\text{cm}}^2\cdot \text{s}}}{\frac{4}{3}\pi {(44)}^3}=27.5\frac{\gamma /\text{sec}}{{\text{cm}}^3}=27.5\frac{\text{Bq}}{{\text{cm}}^{\text{3}}}$

  $C=27.5\times {10}^{-6}\frac{\text{MBq}}{{\text{cm}}^3}$

The source strength

  $\Gamma =3.69\times {10}^{-9}{\displaystyle \sum f_i}{E}_i\frac{(\text{C}/\text{kg}){\text{m}}^2}{\text{MBq}\cdot \text{h}}\times 34\frac{\text{Gy}}{\text{C}/\text{kg}}$

  $\Gamma =3.69\times {10}^{-9}\times 2.26\text{MeV}\frac{(\text{C}/\text{kg}){\text{m}}^2}{\text{MBq}\cdot \text{h}}\times {10}^4\frac{{\text{cm}}^2}{{\text{m}}^2}\times 34\frac{\text{Gy}}{\text{C}/\text{kg}}$

  $\Gamma =2.8\times {10}^{-3}\frac{\text{Gy}\cdot {\text{cm}}^2}{\text{MBq}\cdot \text{h}}=2.8\frac{\text{mGy}\cdot {\text{cm}}^2}{\text{MBq}\cdot \text{h}}$

  ${\mu }_m$ (2.26MeV, H₂0) = 0.0252 cm²/g. Since $H_{2}C\cong H_{2}0$ for gamma energy absorption, we

Linear energy absorption coefficient for paraffin

  ${\mu }_1={\mu }_{\text{m}}\times \rho$${\mu }_1=0.252\frac{{\text{cm}}^2}{g}\times 0.89\frac{\text{g}}{{\text{cm}}^{\text{3}}}=0.022{\text{cm}}^{-1}$

Inserting these values into the dose rate equation, we have

  $\dot{D}=\frac{1}{2}C\Gamma \frac{4\pi }{\mu }\left(1-e^{-\mu r}\right)$

  $\dot{D}=\frac{1}{2}\times \frac{27.5\times {10}^{-6}\text{MBq}}{{\text{cm}}^3}\times \frac{2.8{\text{mGy}}^{\prime }{\text{cm}}^2}{\text{hr}\cdot \text{MBq}}\times \frac{4\times \pi }{0.022{\text{cm}}^{-1}}\left(1-{\text{e}}^{-0.022\times (44)}\right)$

  $\dot{D}=0.014\frac{\text{mGy}}{\text{hr}}=1.4\frac{\text{mrads}}{\text{hr}}$

Conclusion:

Gamma ray dose rate is $1.4\frac{\text{mrads}}{\text{hr}}$