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On bombarding U²³⁵ by slow neutron, 200 MeV energy is released. If the power output of atomic reactor is 1.6 MW, then the rate of fission will be
Options
(a) 8×10¹⁶/s
(b) 20×10¹⁶/s
(c) 5×10²²/s
(d) 5×10¹⁶/s
Correct Answer:
5×10¹⁶/s
Explanation:
Energy released per fission of uranium = 200 × 10⁶ × 1 × 10⁻¹⁹ J
Power output = 1.6 × 10⁶ W
Number of fission /s = 1.6 × 10⁶ / 200 × 10⁶ × 1 × 10⁻¹⁹ = 5 × 10¹⁶ /s
This is the rate of fission.
Related Questions: - Rutherford’s α-scattering experiment concludes that
- In R-L-C series circuit, the potential differences across each element is 20 V.
- Two point objects of masses 1.5 g and 2.5 g respectively are at a distance of 16 cm
- The value of coefficient of volume expansion of glycerin is 5 x 10⁻⁴ k⁻¹.
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Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Rutherford’s α-scattering experiment concludes that
- In R-L-C series circuit, the potential differences across each element is 20 V.
- Two point objects of masses 1.5 g and 2.5 g respectively are at a distance of 16 cm
- The value of coefficient of volume expansion of glycerin is 5 x 10⁻⁴ k⁻¹.
- A block of mass m is moving on a wedge with the acceleration aᵒ
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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