Mean free path `(lambda)` in a fission reaction is the average distance covered by a neutron between two fissions. Generation of nearly `200` neutrons are needed to fission all the nuclei in `10 kg` of `._(92)U^(235)`. If `lambda= 4cm` and the fastest moving neutrons generated in a fission reaction have an energy of `2 MeV`, then the time for `200` neutron generation is found to be `pxx10^(-7)` seconds. Find `'p'`.

Answer

4

Explanation

`v=sqrt((2E)/(m))=2xx10^(7)m//s, t=(lambda)/(v)=2xx10^(-9)s`
For 200 neutrons generation , `T=200 t`
`=4xx10^(-7)s`