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 $1U = 931 \,MeV)$
- $8.5 \,MeV$
- $7.5 \,MeV$
- $8.0\, MeV$
- $9.0\, MeV$
The Correct Option is A
Solution and Explanation
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): \(8.5 \;MeV\)
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In the year 1911, Rutherford discovered the atomic nucleus along with his associates. It is already known that every atom is manufactured of positive charge and mass in the form of a nucleus that is concentrated at the center of the atom. More than 99.9% of the mass of an atom is located in the nucleus. Additionally, the size of the atom is of the order of 10-10 m and that of the nucleus is of the order of 10-15 m.
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