Question

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 $1U = 931 \,MeV)$

• A. $8.5 \,MeV$
• B. $7.5 \,MeV$
• C. $8.0\, MeV$
• D. $9.0\, MeV$

Answer 1

Answer: (A)

$8.5 \,MeV$

Answer 2

B.E. $=[\Delta m ] \cdot c ^{2}$

$M _{\text {expected }} = ZM _{ p }+( A - Z ) M _{ n }$

$=50[1.00783]+70[1.00867]$

$M _{\text {actual }}= 119.902199$

$B.E. =[50[1.00783]+70[1.00867]-119.902199] \times 931$

$= 1020.56$

$\frac{ BE }{\text { nucleon }} =\frac{1020.56}{120}$

$=8.5 \,MeV$

The Correct Option is (A): $8.5 \;MeV$