| ⇦ | 
| ⇨ |
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
Related Questions: - White light reflected from a soap film (Refractive Index =1.5) has a maxima
- An α-particle and a proton are accelerated through same potential difference
- If a carnot engine is working with source temperature at 227⁰C and sink
- A force of 10 newton acts on a body of mass 20 kg for 10 seconds. Change in its momentum
- 24 cells of emf 1.5 V each having internal resistance of 1 ohm are connected
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- White light reflected from a soap film (Refractive Index =1.5) has a maxima
- An α-particle and a proton are accelerated through same potential difference
- If a carnot engine is working with source temperature at 227⁰C and sink
- A force of 10 newton acts on a body of mass 20 kg for 10 seconds. Change in its momentum
- 24 cells of emf 1.5 V each having internal resistance of 1 ohm are connected
Topics: Radioactivity (83)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply