Chapter 10, Problem 10.21P

To determine

To Calculate: The shielding requirements.

Answer to Problem 10.21P

Shielding requirements are:

  $polyethylene=1.1cm$

  $lead=5cm$

Explanation of Solution

Given:

Dose equivalent rate at the surface = $2mSV(200mrems)$

Assumption:

Y-90 is in equilibrium with the Sr-90. Y-90 emits 2.27MeV beta

Useful data:

  $\begin{array}{l}Densityofpolyethylene=0.95g/c{m}^3\\ Formulaforpolyethylene=C{H}_2\\ {\mu }_{en}(muscle,2.27MeV)=0.0249c{m}^2/g\\ \mu (Pb,2.27MeV)=0.506c{m}^{-1}\end{array}$

Formula used:

Range of beta given by, $R=412\times {(E)}^{1.265-0.0954\ln (E)}$

Z_effective is given by

  

Estimated bremsstrahlung production,

  $F_{Y-90}=3.5\times {10}^{-4}ZE$

Beta energy converted to x-ray

  $E_X=F\times E_B$

Calculation:

Range of beta can be calculated by

  $\begin{array}{l}$ R=412\times {(E)}^{1.265-0.0954\ln (E)}$=412\times {(2.27)}^{1.265-0.0954\ln (2.27)}\\ =1090mg/c{m}^2\\ =1.09g/c{m}^2\end{array}$

Thickness of polyethylene required to shield the Y-90 betas can be calculated as follows:

  $=\frac{1.09}{0.95}=1.1cm$

Using the formula of polyethylene

  $\begin{array}{l}{N}_H=\frac{1\times 0.95\times 6.022\times {10}^{23}\times 2}{14}\\ =8\times {10}^{22}atoms/c{m}^3\\ N_C=\frac{1\times 0.95\times 6.022\times {10}^{23}\times 1}{14}\\ =4\times {10}^{22}atoms/c{m}^3\end{array}$

Therefore, $Z_{effective}$ is

  $\begin{array}{l}{Z}_{effective}=\frac{N_1{Z}_1^2+N_2{Z}_2^2}{N_1{Z}_1+N_2{Z}_2}\\ =\frac{(8\times {10}^{22}\times 1^2)+(4\times {10}^{22}\times 6^2)}{(8\times {10}^{22}\times 1)+(4\times {10}^{22}\times 6)}\\ =4.75\end{array}$

Estimated bremsstrahlung production

  $\begin{array}{l}{F}_{Y-90}=3.5\times {10}^{-4}ZE\\ =3.5\times {10}^{-4}(4.75)\times (2.27)\\ =3.8\times {10}^{-3}MeV\end{array}$

E_B is calculated as (using the average energy of betas from the Sr-Y

  $\begin{array}{l}{E}_B=1\times {10}^{11}\times 1.11\\ =1.11\times {10}^{11}MeV/s\end{array}$

Therefore,

  $\begin{array}{l}{E}_X=F\times E_B\\ =3.8\times {10}^{-3}\times 1.11\times {10}^{11}\times \frac{MeV}{s}\times 1.6\times {10}^{-13}\frac{J}{MeV}\\ =6.75\times {10}^{-5}J/s\end{array}$

If the source is surrounded by 1.1 cm plastic, x-ray energy flux of the plastic can be calculated as,

  $\begin{array}{l}{\varphi }_{XE}=\frac{6.75\times {10}^{-5}}{4\pi {(1.1)}^2}\\ =4.44\times {10}^{-6}J/c{m}^{2}s\end{array}$

Therefore, the dose rate is

  $\begin{array}{l}\widehat{H}=\frac{{\varphi }_{XE}\times {\mu }_{en}\times 3.6\times {10}^3}{1\times {10}^{-6}}\\ =\frac{4.44\times {10}^{-6}J/c{m}^{2}s\times 0.0249c{m}^2/g\times 3.6\times {10}^3}{1\times {10}^{-6}}\\ =398.1mSv/h\end{array}$

Minimum thickness t can be calculated as,

  $\begin{array}{l}\frac{H}{H_0}={\left(\frac{1.1}{t+1.1}\right)}^2{e}^{-\mu t}\\ \frac{2}{398.1}={\left(\frac{1.1}{t+1.1}\right)}^2{e}^{-\mu t}\\ t=4.2cm\approx 5cm\end{array}$

Conclusion:

Shielding requirements are,

  $polyethylene=1.1cm$

  $lead=5cm$