Question

A nuclear reactor of 40% efficiency has 10¹⁴ decays/second. If the energy obtained per fission is 250 MeV, then power of reactor is :
(a) 2 kW
(b) 4 kW
(c) 1.6 kW
(d) 3.2 kW

Answer 1

(c) 1.6 kW
Per fission received energy = 250 MeV
= 250 × 10⁶ × 1.6 × 10^-19J
Number of fissions per second = 10¹⁴
∴ Per second received energy
= 250 × 10⁶ × 1.6 × 10^-19 × 10¹⁴
As efficiency is 40%, so output power
P = 250 × 10⁶ × 1.6x 10^-5 × 
= 1.6 × 10³ watt = 1.6 kW