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 QuestionOn 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.

AddThis Website Tools
admin:

Related Questions

  1. α-particles, deuterons and protons of same energy are moving in a perpendicular
  2. The oscillating frequency of a cyclotron is 10 MHz. If the radius of its Dees is o.5 m,
  3. A jar is filled with two non-mixing liquids 1 and 2 having densities
  4. For a simple pendulum performing simple harmonic motion, the time period
  5. A mass of 2.0 kg is put on a flat plan attached to a vertical spring fixed on the ground