Question

If 192 MeV of energy is released due to nuclear fission of each nucleus of U-235, what mass of U-235 undergoes fission per hour in a reactor of power 400 MW?

(Given, 1 amu = 1.66 × 10²⁷ kg)

Answer 1

Given Data:

Power of reactor (P) = 400 MW = 400 × 10⁶ J/s

Time interval (t) = 1 hour = 3600 s

Energy released per fission (E₁) = 192 MeV = 192 × 1.6 × 10⁻¹³ J = 3.072 × 10⁻¹¹ J

Mass of one U-235 nucleus (m) = 235 × 1.66 × 10⁻²⁷ kg = 3.901 × 10⁻²⁵ kg

1. Total Energy Required (E):

E = Power × Time

= (400 × 10⁶) × 3600

= 1.44 × 10¹² J

2. Number of Fissions Per Hour (N):

N = `"Total Energy (E)"/("Energy per fission" (E_1))`

= `(1.44 xx 10^12)/(3.072 xx 10^-11)`

= 4.6875 × 10²²

3. Total Mass of U-235 (M):

M = N × m

= (4.6875 × 10²²) × (3.901 × 10⁻²⁵ kg)

= 0.01828 kg

= 18.2 g

The mass of U-235 that undergoes fission per hour is 18.2 g.