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

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.

admin:

Related Questions

  1. For having large magnification power of a compound microscope
  2. A circular disc of radius 0.2 meter is placed in a uniform magnetic field of induction
  3. A beam of electron passes undeflected through mutually perpendicular
  4. The half-life of a radioactive substance is 30 min. The time (in minutes) taken
  5. A source is moving towards an observer with a speed of 20 m/s and having frequency