Question

200 MeV of energy may be obtained on the fission of one atom of the U${}^{235}$isotope. A reactor using U${}^{235}$ as fuel is generating 1000 kW of power. The rate of nuclear fission in the reactor is

• A. 1000$s^{-1}$
• B. $1.38\times {10}^9$$s^{-1}$
• C. $3.125\times {10}^{16}$$s^{-1}$
• D. 931$s^{-1}$

Answer 1

The correct option is C $3.125\times {10}^{16}$$s^{-1}$

The fission of a single $U^{235}$ nucleus releases $200MeV$ of energy

Energy released in the fission is given by the formula

$E=200MeV$

$=200\times {10}^6\times 1.6\times {10}^{-19}$

$=3.2\times {10}^{-11}J$

$P=\frac{nE}{t}$

$\therefore \frac{n}{t}=\frac{P}{E}$

$=\frac{{10}^6}{3.2\times {10}^{-11}}$

$=3.125\times {10}^{16}{s}^{-1}$