# Obtain the binding energy of the nuclei

Question:

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

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

Solution:

Atomic mass of ${ }_{26}^{56} \mathrm{Fe}, m_{1}=55.934939 \mathrm{u}$

${ }_{26}^{56} \mathrm{Fe}$ nucleus has 26 protons and $(56-26)=30$ neutrons

Where,

Mass of a proton, mH = 1.007825 u

Mass of a neutron, mn = 1.008665 u

∴Δ= 26 × 1.007825 + 30 × 1.008665 − 55.934939

= 26.20345 + 30.25995 − 55.934939

= 0.528461 u

But 1 u = 931.5 MeV/c2

∴Δ= 0.528461 × 931.5 MeV/c2

The binding energy of this nucleus is given as:

Eb1 = Δmc2

Where,

c = Speed of light

$\therefore E_{b 1}=0.528461 \times 931.5\left(\frac{\mathrm{MeV}}{c^{2}}\right) \times c^{2}$

= 492.26 MeV

Average binding energy per nucleon $=\frac{492.26}{56}=8.79 \mathrm{MeV}$

Atomic mass of ${ }_{83}^{209} \mathrm{Bi}, m_{2}=208.980388 \mathrm{u}$

${ }_{83}^{209}$ Bi nucleus has 83 protons and $(209-83) 126$ neutrons.

Hence, the mass defect of this nucleus is given as:

Δm' = 83 × mH + 126 × mn − m2

Where,

Mass of a proton, mH = 1.007825 u

Mass of a neutron, mn = 1.008665 u

∴Δm' = 83 × 1.007825 + 126 × 1.008665 − 208.980388

= 83.649475 + 127.091790 − 208.980388

= 1.760877 u

But 1 u = 931.5 MeV/c2

∴Δm' = 1.760877 × 931.5 MeV/c2

Hence, the binding energy of this nucleus is given as:

$E_{b 2}=\Delta m^{\prime} c^{2}$

$=1.760877 \times 931.5\left(\frac{\mathrm{MeV}}{c^{2}}\right) \times c^{2}$

= 1640.26 MeV

Average bindingenergy per nucleon $=\frac{1640.26}{209}=7.848 \mathrm{MeV}$

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