Question

How many mean free paths thick must a shield be designed in order to attenuate an incident neutron beam by a factor of 1000?

Answer 1

Answer:

A shield should be about 6.91 mean free paths thick to attenuate a neutron beam by a factor of 1000.

Explanation:

The equation for the attenuation of a neutron beam is

$I=I_0 \exp (-x/\lambda)$

where Io, x and λ are the initial intensity, distance traveled and the mean free path of the neutron respectively. I is the intensity after the distance x. Rearranging we get,

$\frac{I}{I_0}=exp(-z/A)$

⇒$e^{-\frac{x}{\lambda} }=\frac{I_0}{I}$

$\frac{x}{\lambda}=ln\frac{I_0}{I}$

$x=\lambda ln\frac{I_0}{I}$

we know that $\frac{I_o}{I}$ = 1000

putting it above equation we get

x=λ ln(1000)

x= 6.91λ

Therefore,  A shield should be about 6.91 mean free paths thick to attenuate a neutron beam by a factor of 1000.