On bombarding U²³⁵ by slow neutron, 200 MeV energy is released. If the power output

On Bombarding U By Slow Neutron 200 Mev Energy Is Physics Question

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:

  1. If the rise in height of capillary of two tubes is 6.6 cm and 2.2 cm
  2. In n-type semiconductor, electrons are majority charge carriers
  3. The frequency of a light wave in a material is 2 x 10¹⁴ Hz and wavelength is 5000Å.
  4. The input characteristics of a transistor in CE mode is the graph obtained by plotting
  5. A particle moves along the x-axis from x=0 to x=5 m under the influence of a force

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.

NEETLab Mobile App

Share this page with your friends

Be the first to comment

Leave a Reply

Your email address will not be published.


*