ExamBro
ExamBro
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

  1. A \(32 \times 10^8\; W\)
  2. B \(16 \times 10^8 \; W\)
  3. C \(5 \times 10^{11}\; W\)
  4. D \(2 \times 10^{22} \; W\)
Verified Solution

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
From NEET
Explore more questions on app