Question

On bombardment of ${}_{92}{U}^{235}$ by slow neutrons, $200MeV$ energy is released. If the power output of atomic reactor is $1.6MW$, then the rate of fission will be:

• A. 5 x10${}^{16}$ / Sec
• B. 10x 10${}^{16}$ / Sec
• C. 15 x10${}^{16}$ / Sec
• D. 20 x10${}^{16}$ / Sec

Answer 1

The correct option is A 5 x10${}^{16}$ / Sec

Energy released in fission reaction $=200MeV$
Power output of the reactor $=1.6MW$
$=1.6\times {10}^{6}watts$

So, Rate of fission $=\frac{1.6\times {10}^6}{200MeV}joules/sec$

$=\frac{1.6}{200\times 1.6\times {10}^{-19}}=5\times {10}^{16}/sec$