In example 31 consider and elastic collision between a neutron and a light nuclei like carbon and calculate fractional KE lost.

Explanation

`f_(1)=((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)` using the result of solved example 31 `m_(2)=12m_(1)`
`f_(1)=((m_(1)-m_(2))/(m_(1)+m_(2)))^(2),` Using the result of solved example 18.
`f=((m_(1)-12m_(2))/(m_(1)+12m_(1)))^(2),` Using `m_(2)=12m_(1)`
`=((-11)/(13))^(2)`
`=121/169`
Expressed as percent it is
`(12100)/(169)=71.6%`
and `f_(2)=28.4%`
In particle, however this number is smaller as head-on collision are none.