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

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

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Bohr's Atomic Model46 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

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

(M = proton mass)

This last expression is not correct because

Physics - Exemplar