Question

The binding energy per nucleon in deuterium and helium nuclei are $1.1 \, MeV$ 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)

$\_{1}^{}H_{}^{2}+\_{1}^{}H_{}^{2} \rightarrow \_{2}^{}He_{}^{4}+ΔE$
The binding energy per nucleon of a deuteron
$=1.1MeV$
∴ Total binding energy $=2\times 1.1=2.2MeV$
The binding energy per nucleon of a helium nuclei
$=7MeV$
∴ Total binding energy $=4\times 7=28MeV$
Hence, the energy released
$ΔE=\left(\right.28-2\times 2.2\left.\right)=23.6MeV$