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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
- Two poles of same strength attract each other with a force of magnitude F
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- Surface tension of a solution is 30 x 10⁻² N/m. The radius of the soap bubble is 5 cm
- The radius of the convex surface of a plano-convex lens is 20 cm and the refractive index
Topics: Radioactivity (83)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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