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Question

Question: The energy liberated on complete fission of 1 kg of \(92U^{235}\) is (Assume 200 MeV energy is liber...

The energy liberated on complete fission of 1 kg of 92U23592U^{235} is (Assume 200 MeV energy is liberated on fission of 1 nucleus).

A

8.2×1010J8.2 \times 10^{10}J

B

8.2×109J8.2 \times 10^{9}J

C

8.2×1013J8.2 \times 10^{13}J

D

8.2×1016J8.2 \times 10^{16}J

Answer

8.2×1013J8.2 \times 10^{13}J

Explanation

Solution

Mass of a uranium nucleus

=92×1.6725×1027+143×1.6747×1027= 92 \times 1.6725 \times 10^{- 27} + 143 \times 1.6747 \times 10^{- 27}

=393.35×1027kg= 393.35 \times 10^{- 27}kg

Number of nuclei in the given mass

=1393.35×1027=2.542×1024= \frac{1}{393.35 \times 10^{- 27}} = 2.542 \times 10^{24}

Energy released =200×2.542×1024MeV= 200 \times 2.542 \times 10^{24}MeV

=5.08×1026MeV=8.135×1013J=8.2×1013J= 5.08 \times 10^{26}MeV = 8.135 \times 10^{13}J = 8.2 \times 10^{13}J