200 MeV of energy may be obtained per fission of U235 . A reactor is generating 1000 kW of power. The rate of nuclear fission in the reaction (in per second) is

Question

200 MeV of energy may be obtained per fission of U235 . A reactor is generating 1000 kW of power. The rate of nuclear fission in the reaction (in per second) is (A)  1000  (B)  2× 108 (C)  3.125× 1016 (D)  931

Tardigrade Admin · Accepted Answer

Energy =200 MeV 200 × 1.6 × 10-19 × 106 =200 × 1.6 × 10-13 Power P =1000 kW =1000 × 103 W Rate of fission =( text Power / text Energy ) =(106/200 × 1.6 × 10-13) =3.125 × 1016 / s

Q. $200 M e V$ of energy may be obtained per fission of $U^{235}$ . A reactor is generating $1000 kW$ of power. The rate of nuclear fission in the reaction (in per second) is

$1000$

$2 \times 1 0^{8}$

$3.125 \times 1 0^{16}$

$931$

Energy $= 200 M e V$
$200 \times 1.6 \times 1 0^{- 19} \times 1 0^{6}$
$= 200 \times 1.6 \times 1 0^{- 13}$
Power $P = 1000 kW$
$= 1000 \times 1 0^{3} W$
Rate of fission $= \frac{Power}{Energy}$
$= \frac{1 0 ^{6}}{200 \times 1.6 \times 1 0 ^{- 13}}$
$= 3.125 \times 1 0^{16} / s$

Q. 200MeV of energy may be obtained per fission of U235 . A reactor is generating 1000kW of power. The rate of nuclear fission in the reaction (in per second) is

1000

2×108

3.125×1016

931