Question

In a reactor, 2 kg of ₉₂U²³⁵ fuel is fully used up in 30 days. The energy released per fission is 200 Me V given that the Avogadro number, N = 6.023 × 10²⁶ per kilo mole and 1 eV = 1.6 × 10^-19 J. The power output of the reactor is close to :

पर्याय

• 54 MW

• 60 MW

• 125 MW

• 35 MW

Answer 1

60 MW

Explanation:

Given: Fuel used in a nuclear reactor is `"U"_"92"^"235"`, mass of the fuel used in t = 30 days is m = 2 kg, energy released per fission reaction is E = 200 Me V, Avogadro's number is N = 6.023 × 10²⁶ per kilo mole, 1 eV = 1.6 × 10^-19 J.

To find: P, power output of the reactor.

Number of `"U"_"92"^"235"`atoms in m = 2 kg of fuel :

n = `(2xx6.023xx10^26)/235`

Energy released on fission of m = 2 kg of fuel :

E' = nE

= `(2xx6.023xx10^26)/235 xx200xx10^6xx1.6xx10^-19"J"`

Power output of the reactor in t = 30 days :

P = `"E'"/"t"`

= `(2xx6.023xx10^26xx200xx10^6xx1.6xx10^-19)/(235xx30xx24xx60xx60)`

P ≈ 60 MW