The binding energy per nucleon of ⁷₃Li and ⁴₂He nuclei are 5.60 MeV and 7.06 MeV respectively. In the nuclear reaction ⁷₃Li + ¹₁H → ⁴₂He + ⁴₂He + Q the value of energy Q released is (a) 19.6 MeV (b) - 2.4 MeV (c) 8.4 MeV (d) 17.3 MeV

Answer: (d). Q = BE of products - BE of reactants (proton has zero BE). BE of 2 ⁴He = 8 × 7.06 = 56.48 MeV; BE of ⁷Li = 7 × 5.60 = 39.20 MeV Q = 56.48 - 39.20 = 17.28 MeV So the answer is (d) 17.3 MeV.

The binding energy per nucleon of ⁷₃Li and ⁴₂He nuclei are 5.60 MeV… | NEET 2014 (AIPMT)

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Nuclear Reactions & Q-value

The binding energy per nucleon of ${}^{7}_{3}Li$ and ${}^{4}_{2}He$ nuclei are 5.60 MeV and 7.06 MeV respectively. In the nuclear reaction

${}^{7}_{3}Li + {}^{1}_{1}H \rightarrow {}^{4}_{2}He + {}^{4}_{2}He + Q$

the value of energy Q released is

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Answer: (d)

$Q$ = BE of products $-$ BE of reactants (proton has zero BE).
BE of $2\,{}^{4}\text{He} = 8 \times 7.06 = 56.48\ \text{MeV}$ ; BE of ${}^{7}\text{Li} = 7 \times 5.60 = 39.20\ \text{MeV}$
$Q = 56.48 - 39.20 = 17.28\ \text{MeV}$
So the answer is (d) $17.3\ \text{MeV}$ .

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