Solveeit Logo
Login

Question

Question: If 200 Me V energy is released in the fission of a single nucleus of \({}_{92}^{235}\)U the fissions...

If 200 Me V energy is released in the fission of a single nucleus of 92235{}_{92}^{235}U the fissions which are required to produce a power of 1 k W is

A

3.125×10133.125 \times 10^{13}

B

1.52×1061.52 \times 10^{6}

C

3.125×10123.125 \times 10^{12}

D

3.125×10143.125 \times 10^{14}

Answer

3.125×10133.125 \times 10^{13}

Explanation

Solution

: Let the number of fission per second be n.

Energy released per second

=n×200MeV=n×200×1.6×10−13J= n \times 200MeV = n \times 200 \times 1.6 \times 10^{- 13}J

Energy required per second= power × time

=1KW×1s=1000J= 1KW \times 1s = 1000J

∴n×200×1.6×10−13=1000\therefore n \times 200 \times 1.6 \times 10^{- 13} = 1000

Or n=10003.2×10−11=103.2×1013=3.125×1013n = \frac{1000}{3.2 \times 10^{- 11}} = \frac{10}{3.2} \times 10^{13} = 3.125 \times 10^{13}