Question

Calculate the binding energy per nucleon of `._(20)^(40)Ca`. Given that mass of `._(20)^(40)Ca` nucleus `= 39.962589 u`, mass of proton `= 1.007825 u`. Mass of Neutron `= 1.008665 u` and `1 u` is equivalent to `931 MeV`.

Answer 1

`A = 40, Z = 20, A-Z = 20`
`Delta m = {Zm_(p) + (A-Z)m_(n)}-M_(n)`
`= {(20 xx 1.007825 +(20 xx 1.008665)} - 39.962589`
`= 40.329800 - 39.962589 , Delta m = 0.367211`
Binding energy per nucleon `=`
`(Delta m xx 931)/(A) = (0.367211 xx 931)/(40) = 8.547 MeV`