Obtain the binding energy of the nuclei ^{56}_{26} Fe \text{ and } ^{209}_{83} Bi in units of MeV from the following data: m ( ^{56}_{26} Fe ) = 55.934939 u, m( ^{209}_{83} Bi ) = 208.980388 u

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Nuclei
... quantities. Therefore, a. Chapter Thirteen. NUCLEI. © NCERT not to be republished ... Chadwick was awarded the 1935 Nobel Prize in Physics for his discovery of ... 13.4 MASS-ENERGY AND NUCLEAR BINDING ENERGY. 13.4.1 Mass – ... of a nucleus and its constituents, ∆M, is called the mass defect, and is given by.
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We express this massive difference in terms of energy by using Einstein's mass energy relationship
Binding energy \Delta E_d = \Delta mc^2
i) ^{56}_{26} Fe
Atomic mass of ^{56}_{26} Fe = 55.934939 amu
number of protons = 26
number of neutrons = 30
Step 1: Finding the mass defect of the given nucleus
Mass of proton m_H = 1.007825 u
Mass of neutron m_n = 1.008665 u
Mass defect of nitrogen \Delta m = m_H + m_n - m
\Delta m=26*(1.007825)+30*(1.008665)-55.934939
\Delta m=0.528741 amu
Converting atomic mass units to MeV/c^2
1 amu = 931.5 MeV/c^2
\Delta m=0.528741*931.5MeV/c^2
\Delta m=492.26\ MeV/c^2
Step 2: Determining binding energy of the given nucleus
Binding energy \Delta E_d = \Delta mc^2
\Delta E_d=492.26MeV/c^2*c^2
\Delta E_d=492.26\ MeV
Therefore, binding energy of the given ^{56}_{26} Fe nucleus \Delta E_d=492.26\ MeV
(ii) ^{209}_{83} Bi
Atomic mass of ^{209}_{83} Bi = 55.934939 amu
number of protons = 83
number of neutrons = 126
Step 1: Finding the mass defect of the given nucleus
\Delta E_d = 104.67 MeV Mass defect of nitrogen \Delta m = m_H + m_n - m
\Delta m=83*(1.007825)+206*(1.008665)-208.980388
\Delta m\ =\ 1.761922\ amu
\Delta m = 1640.26 MeV/c^2
Step 2: Determining binding energy of the given nucleus
\Delta E_d=1640.26\ MeV/c^2*c^2
\Delta E_d=1640.26\ MeV
Hence, binding energy of the given ^{209}_{83} Bi nucleus \Delta E_d=1640.26\ MeV