Q. 44.9( 11 Votes )

Obtain the binding energy of the nuclei 5626Fe and 20983BI in units of MeV from the following data:

m (5626Fe) = 55.934939 u m (20983BI) = 208.980388 u

Answer :

The number of neutron and Proton both in a 56Fe26 atom is respectively 30 and 26.

The mass defect Δm = mp + mn-mFe


And Binding energy is given by EB = Δmc2


Mass of a proton = 1.007825 u


Mass of a neutron = 1.008665 u


In this case mass defect = Δm = (26 × 1.007825 + 30 × 1.008665-55.934939)u = 0.528461u


We know, 1 u = 931.5 MeV/c2


So, E = Δmc2 = 0.528461 × 931.5(MeV/c2) × c2 = 492.26MeV


So, Binding Energy per Nucleon = = = 8.79MeV


The number of neutron and Proton both in a 209Bi83 atom is respectively 126 and 83.


The mass defect Δm = mp + mn-mBi


And Binding energy is given by EB = Δmc2


Mass of a proton = 1.007825 u


Mass of a neutron = 1.008665 u


Mass Defect = Δm = 83 × 1.007825 + 126 × 1.008665-208.980388 = 1.760877 u


We know, 1 u = 931.5 MeV/c2


So, E = Δmc2 = 1.760877 × 931.5 (Mev/c2) × c2 = 1640.26 MeV


Average binding energy per nucleon = 1640.26/209 = 7.848 MeV


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The binding energy of a H-atom, considering an electron moving around a fixed nuclei (proton), is . (m = electron mass). If one decides to work in a frame of reference where the electron is at rest, the proton would be moving around it. By similar arguments, the binding energy would be

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