Question

Find the binding energy of a nucleus cinsisting of equal numbers of protons and neutrons and having the radius one and a half times smaller than that of `Al^(27)` nucleus.

Answer 1

The mass number of the given nucleus must be
`27//((3)/(2))^(3)=8`
Thus the nucleus is `Be^(8)`. Then the binding energy is
`E_(b)= 4xx0.00867+4+x0.00783-0.00531 am u`
`=0.06069am u=56.5MeV`
On using `1 am u= 931 MeV`