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JEE Main Physics Class 12 Atoms and Nuclei Part 2 Questions
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Question : 7
Total: 100
The energy released per fission of nucleus of
240
X
is
200
MeV
. The energy released if all the atoms in
120
g
of pure
240
X
undergo fission is ____
×
10
25
MeV.
(
.
Given
N
A
=
6
×
10
23
)
[24-Jan-2023 Shift 2]
Your Answer:
Validate
Solution:
No. of mole
=
120
240
=
1
2
No. of molecules
=
1
2
×
N
A
Energy released
=
1
2
×
6
×
10
23
×
200
=
6
×
10
25
MeV
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