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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
- In a primary coil 5 A current is flowing on 220 V. In the secondary coil 2200 V
- A common emitter amplifier gives an output of 3 V for an input of 0.01 V.
- Two thin dielectric slabs of dielectric constants K₁ and K₂, (K₁ < K₂) are inserted
- The most important characteristic of electron in the production of X-rays is
- The device can act as a complete electronic circuit is
Topics: Radioactivity (83)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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