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VITEEE 2018 Solved Paper
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© examsnet.com
Question : 29
Total: 125
The Binding energy per nucleon of
7
3
L
i
and
4
2
H
e
nuclei are 5.60 MeV and 7.06 MeV, respectively.
In the nuclear reaction
7
3
L
i
+ H → 2
4
2
H
e
+ Q
the value of energy Q released is :
19.6 MeV
- 2.4 MeV
8.4 MeV
17.3 MeV
Validate
Solution:
BE of
2
H
e
4
= 4 × 7.06 = 28.24 MeV
Be of
7
3
Li = 7 × 5.60 = 39.20 MeV
7
3
Li +
1
1
H
→
2
H
e
4
+
2
H
e
4
+ Q
Therefore, Q = 56.48 - 39.20 = 17.28 MeV.
© examsnet.com
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