Question 6 of 13intermediate🔧 ApplyNumerical3 marks

Obtain the binding energy of the nuclei 2656Fe^{56}_{26}\text{Fe} and 83209Bi^{209}_{83}\text{Bi} in units of MeV from the following data:

m(2656Fe)=55.934939um(^{56}_{26}\text{Fe}) = 55.934939 \text{u} m(83209Bi)=208.980388um(^{209}_{83}\text{Bi}) = 208.980388 \text{u}
Correct Answer

(Bi) Binding Energy of 83209Bi^{209}_{83}\text{Bi} = 1640.26 MeV

(Fe) Binding Energy of 2656Fe^{56}_{26}\text{Fe} = 492.26 MeV

Exercise: EXERCISES | Q: 13.2 | (Page 16)
For More Understanding

Explanation

The binding energy is calculated using the mass defect formula. The mass defect (Δm) is the difference between the sum of masses of individual nucleons and the actual mass of the nucleus. Using the conversion factor 1 u = 931.5 MeV/c² from the context, binding energy = Δm × 931.5 MeV.

Solution Steps

  1. For 2656Fe^{56}_{26}\text{Fe}: Number of protons (Z) = 26, Number of neutrons (A-Z) = 56 - 26 = 30

    Mass of constituent nucleons: = Z × m_H + (A-Z) × m_n = 26×1.00782526 \times 1.007825 + 30×1.00866530 \times 1.008665 = 26.20345 + 30.25995 = 56.46340 u

  2. Mass defect (Δm): Δm = 56.46340 - 55.934939 = 0.528461 u

    Binding Energy: = 0.528461×931.50.528461 \times 931.5 = 492.26 MeV

  3. For 83209Bi^{209}_{83}\text{Bi}: Number of protons (Z) = 83, Number of neutrons (A-Z) = 209 - 83 = 126

    Mass of constituent nucleons: = 83×1.00782583 \times 1.007825 + 126×1.008665126 \times 1.008665 = 83.649475 + 127.09179 = 210.741265 u

  4. Mass defect (Δm): Δm = 210.741265 - 208.980388 = 1.760877 u

    Binding Energy: = 1.760877×931.51.760877 \times 931.5 = 1640.26 MeV

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