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The binding energy per nucleon of $^7_3Li$ and $^4_2He$ nuclei are 5.60 MeV and 7.06 MeV, respectively. In the nuclear reaction $^7_3Li + ^1_1H \rightarrow ^4_2He + ^4_2He + Q$, the value of energy Q released is:
A
19.6 MeV
B
−2.4 MeV
C
8.4 MeV
D
17.3 MeV
Detailed Solution
BE of $^4_2He$ = 4 × 7.06 = 28.24 MeV
BE of $^7_3Li$ = 7 × 5.60 = 39.20 MeV
$Q = 2\times28.24 - 39.20 = 56.48 - 39.20 = 17.28$ MeV ≈ 17.3 MeV
BE of $^7_3Li$ = 7 × 5.60 = 39.20 MeV
$Q = 2\times28.24 - 39.20 = 56.48 - 39.20 = 17.28$ MeV ≈ 17.3 MeV