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Illustrations

Example

Calculate the binding energy of Description: 38906.png assuming the mass of Description: 39051.png atom as 6.01512 amu.

Solution

Mass defect is

 

Δm = [Zmp + (A – Z)mn – M] (1)

 

mp = mass of the proton = 1.007277 amux

 

mn = mass of the neutron = 1.008655 amu

 

Mass of the bound Description: 39055.png = 6.01512 amu

 

(We know that as per the symbol Description: 39059.png, lithium has three protons and three neutrons, since Z = 3, A = 6 and number of neutrons = A – Z = 6 – 3 = 3.)

 

∴ Δm = [3 × 1.007277 + 3(1.008655] – 6.01512

 

= [3.02183 + 3.02596] – 6.01512

 

= 6.047795 – 6.01512 = 0.032675 amu.

 

But the energy equivalent of 1 amu = 931 MeV.

 

Therefore, the binding energy which is equal to the energy equivalent of mass defect in amu is
Eb = Δm × 931 MeV

 

= 0.032675 × 931 = 30.42 MeV

 





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