Question

The binding energy per nucleon of deuteron ${(}_1{H}^2)$ and helium ${(}_{2}H{e}^4)$ are $1.1MeV$ and $7.0MeV$, respectively. The energy released when two deuterons fuse to form a helium nucleus is

• A. $36.2MeV$
• B. $23.6MeV$
• C. $47.2MeV$
• D. $11.8MeV$
• E. $9.31MeV$

Answer 1

The correct option is B $23.6MeV$

The nuclear reaction is,
${}_1{H}^2{+}_1{H}^2{\to }_{2}H{e}^4+Q$
Total binding energy of helium nucleus
$=4\times 7=28MeV$
Total binding energy of each deuteron
$=2\times 1.1=2.2MeV$
Hence, energy released when two deuteron fuse to form helium,
$=28-2\times 2.2=28-44=23.6MeV$.