200 Mev energy is released when one nucleus of $^{235} U$ undergoes fission. Find the number of fissions per second required for producing a power of 1 mega watt

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$3.126 \times 10^{14}$

$3.125 \times 10^{13}$

$3.125 \times 10^{15}$

$3.125 \times 10^{16}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

To calculate the number of fissions per second required to produce a power of 1 mega watt (MW) using the energy released per fission, we can follow these steps:

Convert the power from mega watts to watts.
1 MW = 1,000,000 watts

Convert the energy released per fission from MeV to joules.
$1 M e V = 1.6 x 10^{- 13} j o u l e s$ (approximately)

Calculate the number of fissions per second using the formula:
Power (in watts) = Number of fissions per second × Energy released per fission (in joules)

$1 M W = 1, 000, 000 w a t t s$

Energy released per fission $= 200 M e V = 200 x 1.6 x 10^{- 13} j o u l e s$

Energy released per fission $= 3.2 x 10^{- 11} j o u l e s$

Number of fissions per second = Power (in watts) / Energy released per fission (in joules)
Number of fissions per second $= \frac{1, 000, 000 w a t t s}{3.2 x 10^{- 11} j o u l e s}$