If the binding energy per nucleon in ⁷₃Li and ⁴₂He nuclei are 5.60 MeV and 7.06 MeV respectively, then in the reaction p + ⁷₃Li → 2 ⁴₂He energy of proton must be (a) 28.24 MeV (b) 17.28 MeV (c) 1.46 MeV (d) 39.2 MeV

Answer: (b). Energy released = 8(7.06) - 7(5.60) = 56.48 - 39.20 = 17.28 MeV. The intended answer is this Q value of the reaction. So the answer is (b) 17.28 MeV.

If the binding energy per nucleon in ⁷₃Li and ⁴₂He nuclei are 5.60… | JEE Main 2006 (AIEEE)

Q22JEE Main 2006 · AIEEEMedium

Nuclear Reactions & Q-value

If the binding energy per nucleon in ${}^{7}_{3}Li$ and ${}^{4}_{2}He$ nuclei are $5.60\text{ MeV}$ and $7.06\text{ MeV}$ respectively, then in the reaction

$p + {}^{7}_{3}Li \rightarrow 2\,{}^{4}_{2}He$

energy of proton must be

Show answer and solution

Answer: (b)

Energy released $= 8(7.06) - 7(5.60) = 56.48 - 39.20 = 17.28\ \text{MeV}$ .
The intended answer is this $Q$ value of the reaction.
So the answer is (b) $17.28\ \text{MeV}$ .

Report a mistake

Nuclear Reactions & Q-value question from JEE Main 2006 (AIEEE)

More questions on Nuclear Reactions & Q-value