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Zorluk: OrtaMass Defect and Binding Energy

The atomic mass of a lithium nucleus 37Li^{7}_{3}\text{Li} is 7.0160 u7.0160\text{ u}. Given that the mass of a proton is 1.0073 u1.0073\text{ u} and the mass of a neutron is 1.0087 u1.0087\text{ u}, what is the total binding energy of the lithium nucleus? (1 u=931 MeV1\text{ u} = 931\text{ MeV})

  1. 37.89 MeV37.89\text{ MeV}Cevap
  2. B
    5.41 MeV5.41\text{ MeV}
  3. C
    36.59 MeV36.59\text{ MeV}
  4. D
    0.041 MeV0.041\text{ MeV}

Cevap

The total binding energy of the lithium nucleus is 37.89 MeV37.89\text{ MeV}.
The correct answer is derived by finding the mass defect Δm=[3(1.0073)+4(1.0087)]−7.0160=0.0407 u\Delta m = [3(1.0073) + 4(1.0087)] - 7.0160 = 0.0407\text{ u} and multiplying by 931 MeV/u931\text{ MeV/u} to obtain 37.89 MeV37.89\text{ MeV}.

Adım Adım Çözüm

1
Determine the number of protons and neutrons in 37Li^{7}_{3}\text{Li}
Z=3Z = 3 protons, N=A−Z=7−3=4N = A - Z = 7 - 3 = 4 neutrons
The mass number A=7A=7 and atomic number Z=3Z=3 define the nuclear composition.
2
Calculate total mass of individual nucleons
Mass of nucleons =3(1.0073 u)+4(1.0087 u)=3.0219 u+4.0348 u=7.0567 u= 3(1.0073\text{ u}) + 4(1.0087\text{ u}) = 3.0219\text{ u} + 4.0348\text{ u} = 7.0567\text{ u}
Summing the masses of constituent protons and neutrons.
3
Calculate mass defect (Δm)(\Delta m)
\Delta m = 7.0567\text{ u} - 7.0160\text{ u} = 0.0407\text{ u}
Mass defect is the difference between total constituent mass and nuclear mass.
4
Convert mass defect to energy
E_b = 0.0407\text{ u} \times 931\text{ MeV/u} = 37.8917\text{ MeV} \approx 37.89\text{ MeV}
Using the equivalence 1 u=931 MeV1\text{ u} = 931\text{ MeV}.

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