Question

If the mass of proton = 1.008
a.M.U. And mass of neutron = 1.009
a.M.U. Then binding energy per nucleon for 4be94be9 ( mass = 9.012 amu) would be

Answer 1

Mass defect is -

Δm=[Zmp+(A−Z)mn−M]

Therefore:

Δm=4Mp +(9−4)Mn−MBe

 =(4×1.008+5×1.009−9.012)Amu

=0.065Amu

So Binding energy of  Be = Δm×931.5Mev

=.065×931.5Mev

=60.5475Mev

Or the binding energy per nucleon will be : Be/Nucleon

= 60.5475/9

=6.7275 Mev

Therefore the binding energy per nucleon for 4Be9 is 6.725 Mev