200 MeV of energy may be obtained per fi

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Target Exam

CUET

Subject

Physics

Chapter

Nuclei

Question:

200 MeV of energy may be obtained per fission of U²³⁵. A reactor is generating 1000 kW of power. The rate of nuclear fission in the reactor is

Options:

1000

2 × 10⁸

3.125 × 10¹⁶

931

Correct Answer:

3.125 × 10¹⁶

Explanation:

Power = 1000 kW = $10^6$ J/s

Rate of nuclear fission $=\frac{10^6}{200 \times 1.6 \times 10^{-13}}$ = 3.125 × 10¹⁶