| ⇦ | 
| ⇨ |
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:
- The ratio of minimum wavelengths of Lyman and Balmer series will be
- The angular speed of earth, so that the object on equator may appear weightless
- A radioactive substance contains 10000 nuclei and its halflife period is 20 days
- The lowest frequency of light that will cause the emission of photoelectrons
- The wing span of an aeroplane is 20 m. It is flying in a field where the vertical
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