#### Question

A town has a population of 1 million. The average electric power needed per person is 300 W. A reactor is to be designed to supply power to this town. The efficiency with which thermal power is converted into electric power is aimed at 25%. (a) Assuming 200 MeV to thermal energy to come from each fission event on an average, find the number of events that should take place every day. (b) Assuming the fission to take place largely through ^{235}U, at what rate will the amount of ^{235}U decrease? Express your answer in kg per day. (c) Assuming that uranium enriched to 3% in ^{235}U will be used, how much uranium is needed per month (30 days)?

#### Solution

(a) Total population of the town = 1 million = 10^{6 }

Average electric power needed per person = 300 W

Total power used by the town in one day = 300 × 10^{6} × 60 × 60 × 24 J = 300 × 86400 ×10^{6} J

Energy generated in one fission = 200 × 10^{6} × 1.6 × 10^{−19} J =3.2 × 10^{−11} J

The efficiency with which thermal power is converted into electric power is 25%.

Therefore, Electrical energy is given by

`therefore "Electrical energy" ,"E" = 3.2 xx 10^-11 xx 25/100`

`"E" = 8 xx 10^-12 "J"`

Let the number of fission be N.

So, total energy of N fissions = N × 8 × 10^{−12}

As per the question,

N × 8 × 10^{−12} = 300 × 86400 × 10^{6} J

N = 3.24 × 10^{24}

(b) Number of moles required per day n = `N/(6.023 xx 10^23)`

⇒ `n = (3.24 xx 10^24)/(6.023 xx 10^23) = 5.38 "mol"`

So, the amount of uranium required per day = 5.38 × 235

= 1264.3 gm = 1.2643 kg

(c) Total uranium needed per month = 1.264 × 30 kg

Let x kg of uranium enriched to 3% in ^{235}U be used.

⇒ `x xx 3/100 = 1.264 xx 30`

⇒ `x = 1264 "kg"`