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

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

Step-by-step 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,

`(I)/(I_0)=exp(-z/A)`

⇒
`e^{-(x)/(\lambda) }=(I_0)/(I)`

`(x)/(\lambda)=ln(I_0)/(I)`

`x=\lambda ln(I_0)/(I)`

we know that
`(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.