# A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much ""_92^235"U" did it contain initially? - Physics

Numerical

A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much ""_92^235"U" did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of ""92^235"U" and that this nuclide is consumed only by the fission process.

#### Solution

Half life of the fuel of the fission reactor,  "t"_(1/2) =  years

= 5 × 365 × 24 × 60 × 60 s

We know that in the fission of 1 g of ""_92^235 "U" nucleus, the energy released is equal to 200 MeV.

1 mole, i.e., 235 g of ""_92^235"U"  contains 6.023 × 1023 atoms.

∴1 g ""_92^235 "U" contains (6.023 xx 10^23)/235

The total energy generated per gram of ""_92^235 "U" is calculated as:

"E" = (6.023 xx 10^23)/235 xx 200 " MeV/g"

= (200 xx 6.023 xx 10^23 xx 1.6 xx 10^(-19) xx 10^6)/235

= 8.20 × 1010 J/g

The reactor operates only 80% of the time.

Hence, the amount of ""_92^235"U" consumed in 5 years by the 1000 MW fission reactor is calculated as:

= (5 xx 80 xx 60 xx 60 xx  365 xx 24 xx 1000 xx 10^6)/(100 xx 8.20 xx 10^10)

~~ 1538 "kg"

∴ Initial amount of ""_92^235"U" = 2 × 1538 = 3076 kg

Concept: Nuclear Energy - Nuclear Fission
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Chapter 13: Nuclei - Exercise [Page 464]

#### APPEARS IN

NCERT Physics Class 12
Chapter 13 Nuclei
Exercise | Q 13.18 | Page 464
NCERT Physics Class 12
Chapter 13 Nuclei
Exercise | Q 18 | Page 464
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