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NEET · Physics · STD 12 - 13. Nuclei
The energy liberated per nuclear fission is \(200 \;MeV\). If \(10^{20}\) fissions occur per/second the amount of power produced will be
- A \(32 \times 10^8\; W\)
- B \(16 \times 10^8 \; W\)
- C \(5 \times 10^{11}\; W\)
- D \(2 \times 10^{22} \; W\)
Answer & Solution
Correct Answer
(A) \(32 \times 10^8\; W\)
Step-by-step Solution
Detailed explanation
nuclea fission \(=200 MeV=200 \times 10^{6} \times 1.6 \times 10^{-19} J\)
\(=3.2 \times 10^{-11} J\) Then for \(10^{20}\) fisssion/sec
\(=3.2 \times 10^{-11} J\) Then for \(10^{20}\) fisssion/sec
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