# 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. - Physics

Numerical

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.

#### Solution

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.

Concept: Nuclear Energy - Nuclear Fission
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Chapter 13: Nuclei - Exercise [Page 466]

#### APPEARS IN

NCERT Physics Class 12
Chapter 13 Nuclei
Exercise | Q 13.31 | Page 466
NCERT Physics Class 12
Chapter 13 Nuclei
Exercise | Q 31 | Page 466
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