Question

Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of ²³⁵U to be about 200MeV.

Answer 1

Amount of electric power to be generated, P = 2 × 10⁵ MW

10% of this amount has to be obtained from nuclear power plants.

∴Amount of nuclear power, `P_1 = 10/100 xx 2 xx 10^5`

= 2 × 10⁴ MW

= 2 × 10⁴ × 10⁶ J/s

= 2 × 10¹⁰ × 60 × 60 × 24 × 365 J/y

Heat energy released per fission of a ²³⁵U nucleus, E = 200 MeV

Efficiency of a reactor = 25%

Hence, the amount of energy converted into the electrical energy per fission is calculated 

`25/100 xx 200 = 50 MeV`

` = 50 xx 1.6 xx 10^(-19) xx 10^(6) = 8 xx 10^(-12) J`

Number of atoms required for fission per year:

`(2 xx 10^10 xx 60 xx 60 xx 24 xx 365)/(8 xx 10^(-12)) = 78840 xx 10^(24) "atoms"`

1 mole, i.e., 235 g of U²³⁵ contains 6.023 × 10²³ atoms.

∴ Mass of 6.023 × 10²³ atoms of U²³⁵ = 235 g = 235 × 10⁻³ kg

∴ Mass of 78840 × 10²⁴ atoms of U²³⁵

`= (235 xx 10^(-3))/(6.023 xx 10^23)   xx 78840 xx 10^24`

`= 3.076 xx 10^4 kg`

Hence, the mass of uranium needed per year is 3.076 × 10⁴ kg.