1.

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»

Solution :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`


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