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Q.
Find the binding energy per nucleon for ${ }_{50}^{120} Sn$. Mass of proton $m _{ p }=1.00783\, U ,$ mass of neutron $m_{n}=1.00867 \,U$ and mass of tin nucleus $m _{ Sn }=119.902199 \,U $ (take $1U = 931 \,MeV)$
B.E. $=[\Delta m ] \cdot c ^{2}$
$M _{\text {expected }} = ZM _{ p }+( A - Z ) M _{ n } $
$=50[1.00783]+70[1.00867] $
$ M _{\text {actual }}= 119.902199 $
$ B.E. =[50[1.00783]+70[1.00867]-119.902199] \times 931$
$= 1020.56 $
$\frac{ BE }{\text { nucleon }} =\frac{1020.56}{120}$
$=8.5 \,MeV $