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Wave Optics

Question
CBSEENPH12038668
Wired Faculty App

Obtain the binding energy of the nuclei Fe2656 and Bi83209 in units of MeV from the following data:
m(Fe2656) = 55.934939   u;    m(Bi83209) = 208.980388 u

Solution
i) 
F2656 nucleus contains 26 protons and,
Number of neutrons = (56 - 26) = 30 neutrons 

Now,

Mass of 26 protons = 26 x 1.007825 = 26.20345 u 

Mass of 30 neutrons  = 30 x 1.008665 = 30.25995 u 

Total mass of 56 nucleons = 56.46340 u

Mass of  Fe2656 nucleus = 55.934939 u 

∴ Mass defect, ∆m = 56.46340 - 55.934939         = 0.528461 u 

Total binding energy = 0.528461 x 931.5 MeV = 492.26 MeV 

Average binding energy per nucleon = 492.2656 = 8.790 MeV. 

ii) 
Bi20983 nucleus contains 83 protons and, 

Number of neutrons = (209-83) = 126 neutrons. 

Mass of 83 protons = 83 × 1.007825 = 83.649475 amu 

Mass of 126 neutrons = 126 ×1.008665 = 127.091790 amu 

Therefore, total mass of nucleons = (83.649475+127.091790) = 210.741260 amu 

mass of nucleus = 208.980388 a.m.u (given) 

Now, mass defect, ∆m =  210.741260 - 208.980388 = 1.760872  

Total binding energy = 1.760872 × 931.5 = 1640.26 MeV 

Therefore, average B.E. per nucleon = 1640.26209 = 7.848 MeV 

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