# The fission properties of ""_94^239"Pu" are very similar to those of ""_92^235 "U". The average energy released per fission is 180 MeV. - Physics

Numerical

The fission properties of ""_94^239"Pu" are very similar to those of ""_92^235 "U". The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure ""_94^239 "Pu" undergo fission?

#### Solution

Average energy released per fission of ""_94^239 "Pu", "E"_"av" = 180 "MeV"

Amount of pure ""_94"Pu"^239 m = 1 kg = 1000 g

NA= Avogadro number = 6.023 × 1023

Mass number of ""_94^239 "Pu"= 239 g

1 mole of ""_94"Pu"^239  contains NA atoms.

∴ mg of ""_94"Pu"^239 contain ("N"_"A"/"Mass number" xx m) atom

= (6.023 xx 10^23)/239 xx 1000  = 2.52 xx 10^24 atoms

∴ Total energy released during the fission of 1 kg of ""_94^239"Pu" is calculated as:

"E" = "E"_"av" xx 2.52 xx 10^24

= 180 xx 2.52 xx 10^24 = 4.536 xx 10^26 MeV

Hence, 4.536 xx 10^26 is released if all the atoms in 1 kg of pure ""_94"Pu"^239 undergo fission.

Concept: Nuclear Energy - Nuclear Fission
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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 13 Nuclei
Exercise | Q 13.17 | Page 463
NCERT Class 12 Physics Textbook
Chapter 13 Nuclei
Exercise | Q 17 | Page 463
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