Q: How much (U^{235}) is consumed in a day in an atomic power house operating at 400 MW, provided the

whole of mass U^{235} is converted into energy ?

Sol: Powder = 400 MW = 400 x 10^{6} W;

time = 1 day = 86, 400 s.

Energy produced , E = powder × time = 400 × 10^{6} × 86,400 = 3.456 × 10^{13} J.

As the whole of mass is converted into energy, by Einstein’s mass-energy relation.

Einstein’s mass – energy relation.

E = m c^{2}

E/c^{2} = (3.456 × 10^{13})/(3 × 10^{8})^{2}

= 3.84 × 10^{-4} kg = 0.384 g