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On bombardment of U^235 by slow neutrons...

On bombardment of `U^235` by slow neutrons, 200 MeV energy is released. If the power output of atomic reactor is 1.6 MW, then the rate of fission will be

A

(a) `5xx10^16` per second

B

(b) `10xx10^16` per second

C

(c) `15xx10^16` per second

D

(d) `20xx10^16` per second

Text Solution

Verified by Experts

The correct Answer is:
A
Let N nuclei decay per second. Then,
`N(200xx1.6xx10^-13)=1.6xx10^6`
Solving we get `N=5xx10^16` per sec
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DC PANDEY-MODERN PHYSICS - 2-Level 1 Objective
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  12. A sample of radioactive substance loses half of its activity in 4 days...

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