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
nucleus has 26 protons and (56 - 26)=30 neutrons
Hence, the mass defect of the nucleus,
where,
Mass of a proton,
Mass of neutron,
Step 2: Find the binding energy of nucleus \({ }_{26}^{56} \mathrm{Fe}\).
The binding energy of this nucleus is given as:
where, c = speed of light
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
nucleus has 83 protons and 126 neutrons.
Hence, the mass defect of this nucleus is given as:
where,
Mass of a proton,
Mass of neutron,
But
Step 4: Find the binding energy of \({ }_{83}^{209} \mathrm{Bi}\) nucleus.
Hence, the binding energy of this nucleus is given by,