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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Concepts Used:
Nuclei
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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Following are the terms related to nucleus:
- Atomic Number
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- Atomic Mass Unit