Question

The binding energies per nucleon for deuterium ${(}_1^2\text{H})$ and helium ${(}_2^4\text{H})$ are 1.1 meV and 7.0 meV respectively. The energy released in meV when two deuterons fuse to form a helium nucleus is ______.

(write the nearest integer value after dividing the answer by 6)

Answer 1

Two deuterons fuse to form a helium nucleus:

$2{(}_1^2\text{H})\to {(}_2^4\text{He})$

The energy released in meV is:

$4\left(7.0meV\right)-4\left(1.1meV\right)=23.6meV$

Divide the answer by 6:

The answer to the nearest integer value is:

$\frac{23.6}{6}=4$