| ⇦ | 
| ⇨ |
On bombarding U²³⁵ by slow neutron, 200 MeV energy is released. If the power output of atomic reactor is 1.6 MW, then the rate of fission will be
Options
(a) 8×10¹⁶/s
(b) 20×10¹⁶/s
(c) 5×10²²/s
(d) 5×10¹⁶/s
Correct Answer:
5×10¹⁶/s
Explanation:
Energy released per fission of uranium = 200 × 10⁶ × 1 × 10⁻¹⁹ J
Power output = 1.6 × 10⁶ W
Number of fission /s = 1.6 × 10⁶ / 200 × 10⁶ × 1 × 10⁻¹⁹ = 5 × 10¹⁶ /s
This is the rate of fission.
Related Questions: - The equivalent resistance of two resistors connected in series is 6 Ω
- A circular disc of radius R and thickness R/6 has moment of inertia I about an axis
- The condition for obtaining secondary maxima in the diffraction pattern
- The horizontal range and the maximum height of a projectile are equal
- If a carnot engine is working with source temperature at 227⁰C and sink
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The equivalent resistance of two resistors connected in series is 6 Ω
- A circular disc of radius R and thickness R/6 has moment of inertia I about an axis
- The condition for obtaining secondary maxima in the diffraction pattern
- The horizontal range and the maximum height of a projectile are equal
- If a carnot engine is working with source temperature at 227⁰C and sink
Topics: Atoms and Nuclei (136)
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