NEET2014Physics-Nuclear Physics
NEET 2014 Physics Nuclear Reactions MCQ Question
Type: MCQ-numerical-Hard-Class 12
The binding energy per nucleon of ³Li₇ and ⁴He₂ nuclei are 5.60 MeV and 7.06 MeV respectively. In the nuclear reaction ³Li₇ + ¹H → ⁴He₂ + ⁴He₂ + Q, the value of energy Q released is
A
19.6 MeV
B
−2.4 MeV
C
8.4 MeV
D
17.3 MeV
Correct Answer
Option D
Detailed Explanation
The total binding energy before the reaction is 3 × 5.60 + 1 × 0 = 16.8 MeV. After the reaction, it is 2 × 4 × 7.06 = 56.48 MeV. The energy released is 56.48 − 16.8 = 39.68 MeV.
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