Question

Obtain the binding energy of the nuclei \(^{56}_{26}Fe\) 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

Answer 1

Atomic mass of  \(^{56}_{26}Fe\) , \(m1 = 55.934939 u\)
\(^{56}_{26}Fe\) nucleus has 26 protons and (56 − 26) = 30 neutrons
Hence, the mass defect of the nucleus, \(∆m = 26 × m_H + 30 × m_n − m_1 \)
Where, 
Mass of a proton, \(m_H = 1.007825 u \)
Mass of a neutron, \(m_n = 1.008665 u \)
∆m = 26 × 1.007825 + 30 × 1.008665 − 55.934939 
= 26.20345 + 30.25995 − 55.934939 
= 0.528461 u 
But 1 u = 931.5 \(\frac{MeV}{c^2 }\)
∆m = 0.528461 × 931.5 \(\frac{MeV}{c^2 }\)
The binding energy of this nucleus is given as:
\(E_{b1} = ∆mc^2 \)
Where, c = Speed of light
\(E_{b1}\) = 0.528461 × 931.5 \((\frac{MeV}{c^2 })\times c^2\)
\(E_{b1}\) = 492.26 MeV
Average binding energy per nucleon = \(\frac{492.26}{56} = 8.79 MeV\)

Atomic mass of \( ^{209} _{83} Bi\), \(m_2 = 208.980388 u\)
\( ^{209} _{83} Bi\) nucleus has 83 protons and (209 − 83) 126 neutrons.
Hence, the mass defect of this nucleus is given as: 
\(∆m' = 83 × m_H + 126 × m_n − m_2 \)
Where, 
Mass of a proton, \(m_H = 1.007825 u\) 
Mass of a neutron, \(m_n = 1.008665 u \)
\(∆m' \)= 83 × 1.007825 + 126 × 1.008665 − 208.980388 
\(∆m' = \) 83.649475 + 127.091790 − 208.980388 
\(∆m' = 1.760877 u \)
But \(1 u = 931.5 \frac{MeV}{c^2} \)
∆m' = 1.760877 × 931.5\(\frac{MeV}{c^2} \)
Hence, the binding energy of this nucleus is given as:
\(E_{b2} = ∆m'c^2\)
\(E_{b2}\) = 1.760877 × 931.5 \((\frac{MeV}{c^2})c^2\)
\(E_{b2} \)= 1640.26 MeV
Average binding energy per nucleon =\(\frac{1640.26}{209} =\) 7.848 MeV