The binding energy per nucleon of deuteron $\left({ }_{1}^{2} \mathrm{H}\right)$ and helium nucleus $\left({ }_{2}^{4} \mathrm{He}\right)$ is 1.1 MeV and 7 MeV respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is

JEE Main · 2004 · 24 Apr 2004 · Offline · Modern Physics

The binding energy per nucleon of deuteron

(_{1}^{2} H)

and helium nucleus

(_{2}^{4} He)

is 1.1 MeV and 7 MeV respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is

(A)13.9 MeV
(B)26.9 MeV
(C)23.6 MeV✓ Correct
(D)19.2 MeV

Answer: (C)

Solution

The energy released equals the binding energy of the product less that of the reactants. The helium nucleus carries

4 \times 7 = 28

MeV of binding energy, while the two deuterons together carry

2 \times (2 \times 1.1) = 4.4

MeV. The difference,

28 - 4.4 = 23.6

MeV, is the energy set free.

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option

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