Answer:
The nuclear binding energy = [tex]1.466\times 10^{-13} KJ[/tex]
The binding energy per nucleon =1.37×10⁻¹⁵ KJ/nucleon
Explanation:
Given:
Number of protons = 47
Number of neutrons = 107-47 = 60
Now,
the mass defect (m)= Theoretical mass - actual mass
m = [tex]47 (1.007825) + 60(1.008665) - 106.9051[/tex]
since,
mass of proton = 1.007825 amu
Mass of neutron = 1.008665 amu
thus,
m = [tex]47.367775 + 60.5199 - 106.9051[/tex]
or
m = [tex]0.982575 amu[/tex]
also
1 amu = [tex]1.66\times 10^{-27} kg[/tex]
therefore,
m = [tex]0.982575 amu \times 1.66\times 10^{-27} kg[/tex]
or
m = [tex]1.6310745\times 10^{-27} kg[/tex]
now,
Energy = mass × (speed of light)²
thus,
Energy = [tex]1.6310745\times 10^{27} kg ( 3\times 10^8 ms^{-1} )[/tex]
or
Energy = [tex]14.66\times 10^{-11} kgm^2/s^2[/tex]
or
Energy = [tex]1.466\times 10^{-10} J[/tex] = [tex]1.466\times 10^{-13} KJ[/tex]
Therefore the nuclear binding energy = [tex]1.466\times 10^{-13} KJ[/tex]
Now,
the binding energy per nucleon = [tex]\frac{1.466\times 10^{-13} KJ}{107}[/tex] = 1.37×10⁻¹⁵ KJ/nucleon
