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If the binding energy per nucleon in `L i^7` and `He^4` nuclei are respectively `5.60 MeV` and `7.06 MeV`. Then energy of reaction `L i^7 + p rarr 2_2 He^4` is.
A. `19.6MeV`
B. `2.4MeV`
C. `8.4MeV`
D. `17.3MeV`

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Correct Answer - D
`._(3)Li^(7)+_(1)H^(1)rarr 2_(2)He^(4)`
Energy of reaction `= 8xx7.06 - 7xx5.60 = 17.28MeV`

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