13.4 Obtain the binding energy of the nuclei \({ }_{26}^{56} \mathrm{Fe}\) and \({ }_{83}^{209} \mathrm{Bi}\) in units of MeV from the following data:

m (\({ }_{26}^{56} \mathrm{Fe}\)) = 55.934939 u and m (\({ }_{83}^{209} \mathrm{Bi}\)) = 208.980388 u

Hint: The binding energy can be obtained by finding the mass defect in the nucleus.
Step 1: Find the mass defect in nucleus \({ }_{26}^{56} \mathrm{Fe}\).
Atomic mass of Fe2656, m1=55.934939 u
Fe2656 nucleus has 26 protons and (56 - 26)=30 neutrons
Hence, the mass defect of the nucleus, m=26×mH+30×mn-m1
where,
Mass of a proton, mH=1.007825 u
Mass of neutron, mn=1.008665 um=26×1.007825+30×1.008665-55.934939=26.20345+30.25995-55.934939=0.528461 u
But 1u=931.5 MeV/c2
m=0.528461×931.5 MeV/c2
Step 2: Find the binding energy of nucleus \({ }_{26}^{56} \mathrm{Fe}\).
The binding energy of this nucleus is given as:
Eb1=mc2 where, c = speed of light 
Eb1=0.52861×931.5MeVc2c2=492.400215 MeV
Average binding energy per nucleon=\(\frac{492.400215}{56}=8.792 ~MeV\)
Step 3: Find the mass defect in nucleus \({ }_{83}^{209} \mathrm{Bi}\).
Atomic mass of Bi83209, m2=208.980388 u
Bi83209 nucleus has 83 protons and (209-83) 126 neutrons.
Hence, the mass defect of this nucleus is given as:
m' =83×mH+126×mn-m2
where,
Mass of a proton, mH=1.007825 u
Mass of neutron, mn=1.008665 u
m'=83×1.007825+126×1.008665-208.980388=83.649475+127.091790-208.980399=1.760877 u
But 1 u=931.5 MeV/c2
Step 4: Find the binding energy of  \({ }_{83}^{209} \mathrm{Bi}\) nucleus.
Hence, the binding energy of this nucleus is given by,
Eb2=m'c2
=1.760877×931.5MeVc2c2=1640.26 MeV
Average binding energy per nucleon =1640.26209=7.848 MeV