Question

The binding energy per nucleon of deuterium and helium nuclei are $1.1$ Me V and $7.0$ MeV respectively when two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is.

• A. $23.6MeV$
• B. $2.2MeV$
• C. $28.0MeV$
• D. $30.2MeV$

Answer 1

Answer: (A)

$23.6MeV$

Answer 2

: $1H^{2} +_{1}H^{2} \rightarrow_{2}He^{4} + \Delta E$

The binding energy per nucleon of a deuterium $= 1.1MeV$

$\therefore$Total binging energy

$= 2 \times 1.1 = 2.2MeV$

The binding energy per nucleon of a helium nuclei $= 7MeV$

$\therefore$Total binding energy

$= 4 \times 7 = 28MeV$

Hence, energy released

$\Delta E = (28 - 2 \times 2.2) = 23.6MeV$