Test Index

NCERT Class XII Chapter
Nuclei
Questions With Solutions

© examsnet.com
Question : 17
Total: 31
The fission properties of
239
‌
94
Pu are very similar to those of
235
‌
92
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
239
‌
94
Pu undergo fission?
Solution:  
Number of atoms present in 1 mole i.e.,
239 g of
239
‌
94
Pu = 6.023 × 1023
∴ Number of atoms present in 1000 g of
239
‌
94
Pu
=
6.023×1023×1000
239
 = 2.52 × 1024
Energy released per fission = 180 MeV
Total energy released = 2.52 × 1024 × 180 MeV = 4.54 × 1026 MeV.
© examsnet.com
Go to Question:
Prev Question
Next Question