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