1H^2 + 1H^2 → 2He^4 If binding energy per nucleon of deuterium is 1.1 eV and of helium

Question

\({}_1 H^2 + {}_1H^2 \rightarrow {}_2 He^{4}\)

If binding energy per nucleon of deuterium is 1.1 eV and of helium is 7.0 eV. Find energy released in reaction :-

(1) 2.8 eV

(2) 23.6 eV

(3) 14.8 eV

(4) None

Answer 1

Correct option is : (2) 23.6 eV

\(2_1 H^2 \rightarrow {}_2 He^4\)

\(\Delta E = 4 \times BE\ {}_2 He^4 - (2\times {}_1 \ H^2)\)

\( = (4\times 7.0 - 4\times 1.1)\ eV\)

\( = 4\times 5.4\ MeV = 23.6\ eV\)