If 200 MeV of energy is released in the fission of 1 nucleus of `._(92)U^(235)`, the number of nuclei that undergo fission to produce energy of 10 kWh in 1 s is

Options

A. `11.25xx10^(18)`
B. `22.5xx10^(17)`
C. `11.25xx10^(17)`
D. `22.5xx10^(18)`

Answer

C

Explanation

`(1)/(2)m_(1)u_(1)^(2)-(1)/(2)m_(1)v_(1)^(2)=(75)/(100)xx(1)/(2)m_(1)u_(1)^(2)`
`implies u_(1)^(2)-v_(1)^(2)=(3)/(4)u_(1)^(2) implies v_(1)=(1)/(2)u_(1)` . .. (i)
Now `v_(1)=((m_(2)-m_(1))u_(1))/((m_(1)+m_(2)))` .. . (ii)
thus, `(1)/(2)u_(1)=((m_(2)-m_(1))u_(1))/((m_(1)+m_(2)))impliesm_(2)=3m_(1)=3m`