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

The Correct Option is A

To calculate the binding energy per nucleon for the tin isotope \(\mathrm{{}^{120}_{50}Sn}\), we need to follow these steps:

1. Determine the number of protons and neutrons in the nucleus:
   • The atomic number \(Z\) of tin (\(\mathrm{Sn}\)) is 50, so there are 50 protons.
   • The mass number \(A\) is 120, so the number of neutrons \(N = A - Z = 120 - 50 = 70\) neutrons.
2. Calculate the total mass of the separate nucleons:
   • Mass of 50 protons: \(50 \times 1.00783 \, \text{u} = 50.3915 \, \text{u}\)
   • Mass of 70 neutrons: \(70 \times 1.00867 \, \text{u} = 70.6069 \, \text{u}\)
   • Total mass of separate nucleons: \(50.3915 \, \text{u} + 70.6069 \, \text{u} = 120.9984 \, \text{u}\)
3. Find the actual mass of the nucleus given in the problem:
   • Mass of the tin nucleus (\(m_{\mathrm{Sn}}\)): \(119.902199 \, \text{u}\)
4. Determine the mass defect (\(\Delta m\)):
   • \(\Delta m = \text{Total mass of separate nucleons} - \text{Mass of the nucleus}\)
   • \(\Delta m = 120.9984 \, \text{u} - 119.902199 \, \text{u} = 1.096201 \, \text{u}\)
5. Convert the mass defect to energy using \(1 \, \text{u} = 931 \, \text{MeV}\):
   • Binding energy (\(E_B\)) = \(\Delta m \times 931 \, \text{MeV}\)
   • \(E_B = 1.096201 \, \text{u} \times 931 \, \text{MeV/u} = 1020.423331 \, \text{MeV}\) (approximately)
6. Calculate the binding energy per nucleon:
   • Binding energy per nucleon \(E_{\mathrm{Bn}}\) = \(\frac{E_B}{A}\)
   • \(E_{\mathrm{Bn}} = \frac{1020.423331 \, \text{MeV}}{120}\)
   • \(E_{\mathrm{Bn}} \approx 8.5035 \, \text{MeV}\)

Upon rounding, the binding energy per nucleon is approximately \(8.5 \, \text{MeV}\). Thus, the correct answer is \(8.5 \, \text{MeV}\).