Question 13.30:
Calculate and compare the energy released by a) fusion of 1.0 kg of hydrogen deep within Sun and b) the fission of 1.0 kg of 235U in a fission reactor.
Answer:
(a) Amount of hydrogen, m = 1 kg = 1000 g
1 mole, i.e., 1 g of hydrogen ( ) contains 6.023 × 1023 atoms.
∴1000 g of contains 6.023 × 1023 × 1000 atoms.
Within the sun, four nuclei combine and form one nucleus. In this process 26 MeV of energy is released.
Hence, the energy released from the fusion of 1 kg is:
(b) Amount of = 1 kg = 1000 g
1 mole, i.e., 235 g of contains 6.023 × 1023 atoms.
∴1000 g of contains
It is known that the amount of energy released in the fission of one atom of is 200 MeV.
Hence, energy released from the fission of 1 kg of is:
∴
Therefore, the energy released in the fusion of 1 kg of hydrogen is nearly 8 times the energy released in the fission of 1 kg of uranium.
Question 13.31:
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:
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,
= 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 as:
Number of atoms required for fission per year:
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
Hence, the mass of uranium needed per year is 3.076 × 104 kg.