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.

Related Questions:

1. Mass defect is known as
2. A person can see clearly objects only when they lie between 50 cm and 400 cm from
3. The maximum number of possible interference maxima for slit-seperation equal to
4. An unpolarised beam of intensity I₀ falls on a polaroid. The intensity
5. In an inductor when current changes from 2 A to 18 A in 0.05 sec, the e.m.f. induced

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