JEE MAIN - Physics (2020 - 2nd September Morning Slot - No. 3)
In a reactor, 2 kg of 92U235 fuel is fully used up
in 30 days. The energy released per fission is
200 MeV. Given that the Avogadro number,
N = 6.023 $$ \times $$ 1026 per kilo mole and 1 eV =
1.6 × 10–19 J. The power output of the reactor is
close to
125 MW
60 MW
54 MW
35 MW
Explanation
Number of uranium atoms in 2 kg
= $${{2 \times 6.023 \times {{10}^{26}}} \over {235}}$$
Energy from one atom is 200 × 106 e.V. hence total energy from 2 kg uranium
= $${{2 \times 6.023 \times {{10}^{26}}} \over {235}} \times 200 \times {10^6}$$ eV
= $${{2 \times 6.023 \times {{10}^{26}}} \over {235}} \times 200 \times {10^6}$$ $$ \times $$ 1.6 × 10–19 J
2 kg uranium is used in 30 days hence energy received per second or power is
Power = $${{2 \times 6.023 \times {{10}^{26}} \times 200 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}} \over {235 \times 30 \times 24 \times 3600}}$$
= 63.2 × 106 watt or 63.2 Mega Watt
$$ \simeq $$ 60 MW
= $${{2 \times 6.023 \times {{10}^{26}}} \over {235}}$$
Energy from one atom is 200 × 106 e.V. hence total energy from 2 kg uranium
= $${{2 \times 6.023 \times {{10}^{26}}} \over {235}} \times 200 \times {10^6}$$ eV
= $${{2 \times 6.023 \times {{10}^{26}}} \over {235}} \times 200 \times {10^6}$$ $$ \times $$ 1.6 × 10–19 J
2 kg uranium is used in 30 days hence energy received per second or power is
Power = $${{2 \times 6.023 \times {{10}^{26}} \times 200 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}} \over {235 \times 30 \times 24 \times 3600}}$$
= 63.2 × 106 watt or 63.2 Mega Watt
$$ \simeq $$ 60 MW
Comments (0)
