JEE Main · 2006 · 30 Apr 2006 · Offline · Modern Physics

If the binding energy per nucleon in

_{3}^{7} Li

and

_{2}^{4} He

nuclei are 5.60 MeV and 7.06 MeV respectively, then in the reaction

p +_{3}^{7} Li \to 2_{2}^{4} He

energy of proton must be

(A)39.2 MeV
(B)28.24 MeV
(C)17.28 MeV✓ Correct
(D)1.46 MeV

Answer: (C)

Solution

The energy balance in

p +_{3}^{7} Li \to 2_{2}^{4} He

is governed by the change in total binding energy. The two helium nuclei carry

2 \times (4 \times 7.06) = 56.48 MeV

of binding energy, while the lithium nucleus has

7 \times 5.60 = 39.2 MeV

and the free proton none. The reaction releases the difference,

56.48 - 39.2 = 17.28 MeV

, which is the energy the proton must effectively bring in for the books to balance.

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option

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