If the energy released in the fission of the nucleus is `200 MeV`. Then the number of nuclei required per second in a power plant of `6 kW` will be.

Options

A. `0.5 xx 10^14`
B. `0.5 xx 10^12`
C. `5 xx 10^12`
D. `5 xx 10^14`

Answer

D

Explanation

(d) Energy released in the fission of one nucleus
=`200 MeV`
=`200 xx 10^6 xx 1.6 xx 10^-19 J = 3.2 xx 10^-11 J`
`P = 16 KW = 16 xx 10^3 Watt`
Now, number of nuclei required per second
`n = (P)/( E) = (16 xx 10^3)/(3.2 xx 10^-11) = 5 xx 10^14`.