Question 2.2: The uncollided flux at a distance r from a point source emit......
The uncollided flux at a distance r from a point source emitting is given by Eq. (2.9):
a. If you are 1 m away from a very small 1 Curie source of neutrons. What is the flux of neutrons in n/cm²/s , neglecting scattering and absorption in air.
b. If a shield is placed between you and the source, what absorption cross section would be required to reduce the flux by a factor of 10?
c. Suppose the shield made of the material specified in part b. is only 0.5 m thick. How far must you be from the source, for the flux to be reduced by the same amount as in part b.
a. \quad1 \;\text{Ci} =3.7 \times 10^{10}\, \text{n/s} ;\;\; 1 m = 100 cm
I(100)=\frac{\exp (-0)}{4 \pi 100^2} 3.7 \cdot 10^{10}=294 \cdot 10^3 \;n / cm ^2 \,s
b. \quad\exp (-\Sigma \cdot 100)=1 / 10 \quad Therefore \;\Sigma=\frac{1}{100} \ln (10)=0.023 \,cm ^{-1}
c. \quad \frac{\exp (-\Sigma \cdot 50)}{4 \pi R^2} S=\frac{\exp (-\Sigma \cdot 100)}{4 \pi 100^2} S\quad \text{Hence: }\;R^2=10^4 \exp (\Sigma \cdot 50)
R = 179 cm = 1.79 m