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Q. The mass of helium atom of mass number $4$ is $4.0026$ amu, while that of the neutron and proton are $1.0087$ and $1.0078$ respectively on the same scale. Hence, the nuclear binding energy per nucleon in the helium atom is nearly

Uttarkhand PMTUttarkhand PMT 2008

Solution:

$He$ atom has $2 p+2 n$
Hence, $\Delta m=(2 \times 1.0078+2 \times 1.0087)-4.0026$
$=0.0304 \,a m u$
$\therefore $ energy released $=0.0304 \times 931.5\, MeV$
$=28.3 \,MeV$
Binding energy per nucleon
$=\frac{28.3}{4}=7 \,MeV$ (approximately)