How much U 235 is consumed in a day in an atomic power house operating at 400 M W , provided the whole of the mass of U 235 is converted into energy?

Step-by-step solution

Here, rate of production of energy at the atomic power house, P = 400 M W = 400 × 10 6 J s - 1 Therefore, total energy produced in a day i.e., 24 × 60 × 60 s , E = P × 24 × 60 × 60 = 400 × 10 6 × 24 × 60 × 60 = 3.456 × 10 13 J If mass of U 235 consumed per day in m (in kg) so as to produce the required amount of energy, then E = m c 2 Or 3.456 × 10 13 = m c 2 Or m = 3.456 × 10 13 c 2 = 3.456 × 10 13 3 × 10 8 2 = 0.384 × 10 - 3 k g = 0.384 g