JEE Mains · Physics · STD 12 - 13. Nuclei
Find the bindng energy per nucleon for \(\frac{50}{120}\, Sn\). Mass of proton \(m _{ P }=1.00783\, U ,\) mass of \(m _{ n }=1.00867\, U\) and \(mass\) of tin nucleus \(m _{ Sn }=119.902199\) \(U.\)\(......MeV\) (take \(1 U =931\, MeV )\)
- A \(8.5\)
- B \(7.5\)
- C \(8.0\)
- D \(9.0\)
Answer & Solution
Correct Answer
(A) \(8.5\)
Step-by-step Solution
Detailed explanation
\(B.E =[\Delta m ] \cdot c ^{2}\) \(M _{\text {expected }}= ZM _{ p }+( A - Z ) M _{ n }\) \(=50[1.00783]+70[1.00867]\) \(B E .=[50[1.00783]+70[1.00867]-119.902199]\)\(\times 931\) \(=1020.56\) \(\frac{ BE }{\text { nucleon }}=\frac{1020.56}{120}=8.5 MeV\)
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