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The energy released by the fission of one uranium atom is 200 MeV. The number of fission per second required to produce 6.4 W power is
Options
(a) 2×10¹¹
(b) 10¹¹
(c) 10¹⁰
(d) 2×10¹⁰
Correct Answer:
2×10¹¹
Explanation:
Let there is n number of fission per second produces a power of 6.4 W, then
n × 200 × 10⁶ × 1.6 × 10⁻¹⁹ Js⁻¹ = 6.4 Js⁻¹
.·. n = 6.4 / 200 × 10⁻¹³ × 1.6 = 4 / 2 × 10¹¹
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Topics: Radioactivity
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Subject: Physics
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire
- The intermediate image formed by the objective of a compound microscope is
- The identical loops of copper and aluminium are moving with the same speed
- As a result of interference of two coherent sources of light, energy is
- A steady current of 1.5 amp flows through a copper voltmeter for 10 minutes
Topics: Radioactivity (83)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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