Question

Obtain the binding energy of the nuclei Fe and Bi in units of MeV from the following data. Refer image for question and equation.

Answer 1

we know,
mass of proton ,$m_p$ = 1.00783u
mass of neutron , $m_n$ = 1.00867u

(1) for $^{56}_{26}Fe$,
number of protons = 26
number of neutrons = 30
so, mass defect , ∆m = $(26m_p+30m_n)-m$
= (26 × 1.00783u + 1.00867u) - 55.934939u
= 0.528741u

so, binding energy , E = ∆m × 931MeV
= 0.528741 × 931MeV
= 492.257871MeV


(2) for $^{209}_{83}Bi$
number of protons = 83
number of neutrons = 126

so, mass defect, ∆m = $(83m_p+126m_n)-m$
= (83 × 1.00783u + 126 × 1.00867u) - 208.980388u
= 1.761922u

now, binding energy, E = ∆m × 931MeV
= 1.761922 × 931 MeV
= 1,640.34938MeV