| ⇦ |
| ⇨ |
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 half-life of a radioactive substance is 30 min. The time (in minutes) taken
- A parallel plate capacitor as a uniform electric field E in the space between the plates
- A proton is projected with a speed of 3×10⁶ m/s horizontally from east to west.
- Ist law of thermadynamics is based on
- There are four light-weight-rod samples A,B,C,D seperately suspended by threads
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The half-life of a radioactive substance is 30 min. The time (in minutes) taken
- A parallel plate capacitor as a uniform electric field E in the space between the plates
- A proton is projected with a speed of 3×10⁶ m/s horizontally from east to west.
- Ist law of thermadynamics is based on
- There are four light-weight-rod samples A,B,C,D seperately suspended by threads
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