Question

If 200 MeV energy is released in the fission of a single $U^{235}$ nucleus, the number of fissions required per second to produce 1 kilowatt power shall be (Given 1eV = 1.6 × 10⁻¹⁹ J)(a) 3.125 × 10¹³(b) 3.125 × 10¹⁴(c) 3.125 × 10¹⁵(d) 3.125 × 10¹⁶

Answer 1

To produce 1 kilowatt power number of fissions required per second will be option (a) 3.125 × 10¹³

To produce 1 kW of power for one second, energy required

E = P x t = 1000 x 1 = 1000 joules

Given one fission produces  200 MeV energy

= 200 x 10⁶ x 1.6 × 10⁻¹⁹

= 3.2 x 10⁻¹¹ joules of energy

Hence to produce 1000 joules of energy, number of fission required will be,

n = 1000/(3.2 x 10⁻¹¹) = 3.125 x 10¹³

Hence to produce 1 kilowatt power number of fissions required per second will be option (a) 3.125 × 10¹³

Answer 2

Answer:

(a) 3.125 × 10¹³

Explanation:

Energy = 200 MeV (Given)

Fission nucleus = U^{235} (Given)

Power = 1 kilowatt (Given)

To produce 1 kW of power for one second the energy required will be -

E = P x t ( where p is the power and t is the time)

= 1000 x 1 = 1000 joules

Since, one fission produces 200 MeV energy

= 200 x 10⁶ x 1.6 × 10⁻¹⁹

= 3.2 x 10⁻¹¹ joules of energy

Therefore, for 1000 joules of energy, number of fissions will be,

n = 1000/(3.2 x 10⁻¹¹) = 3.125 x 10¹³

Thus, to produce 1 kilowatt power number of fissions required per second will be option 3.125 × 10¹³.