Question

In a nuclear reactor, $\_{}^{235}U$ undergoes Fission liberating $200 \, MeV$ of energy. The reactor has a $10\%$ efficiency and produces $1000 \, MW$ power. If the reactor is to function for $10 \, yr$ , find the total mass of uranium required.

• A. 38470
• B. 38490
• C. 48490
• D. 48470

Answer 1

Answer: (A)

The reactor produces 1000 MW power or 109J/s. The reactor is to function for 10 yr. Therefore, total energy which the reactor will supply in 10 yr is
E = (power) (time)
= (10⁹J/s) (10 x 365 x 24 x 3600s)
= 3. 1536 x 10¹⁷ J
But since the efficiency of the reactor is only 10%, therefore actual energy needed is 10 times of it or 3. 1536 x $10^{ 18}$ J. One uranium atom liberates 200 MeV of energy or 200 x 1. 6 x $10^{- 13}$ J or 3. 2 x $10^{- 11}$ J of energy. So, number of uranium atoms needed are
$\frac{3.1536 \times 1 0^{18}}{3.2 \times 1 0^{- 11}} = 0.9855 \times 1 0^{29}$
or number of kg - moles of uranium needed are
$n = \frac{0.9855 \times 1 0^{29}}{6.02 \times 1 0^{26}} = 163.7$
Hence, total mass of uranium required is
m = (n)M
or = (163. 7) (235) kg
or m = 38470 kg
or m = 38470 kg