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JEE Mains · Physics · STD 12 - 13. Nuclei

\(\frac{50}{120}\, Sn\) ની ન્યુક્લિયોનદીઠ બંધન ઉર્જા શોધો.( \(MeV\) માં) પ્રોટોનનું દળ \(m _{ P }=1.00783\, U ,\) ન્યૂટ્રોનનું દળ \(m _{ n }=1.00867\, U\) અને ન્યુક્લિયસનું દળ \(m _{ Sn }=119.902199\) \(U.\) (લો : \(1 U =931\, MeV )\)

  1. A \(8.5\)
  2. B \(7.5\)
  3. C \(8.0\)
  4. D \(9.0\)
Verified Solution

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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