Question

What is the binding energy per nucleon in `._2He^4`
Given , Mass of `._2He^4`=4.002604 amu
Mass of proton =1.007825 amu
Mass of neutron =1.008665 amu

Answer 1

No. of protons in ₂He⁴ = 2

No. of neutrons in ₂He⁴ = 2

Mass of 2 protons = 2 x 1.007825 amu

= 2.01565 amu

Mass of 2 neutrons = 2 x 1.008665 amu

= 2.01733 amu

∴ Mass defect, △m = [2.01565 + 2.01733 - 4.002604]

= 0.030376 amu

Binding energy, BE = △m x 931 MeV

= 0.030376 x 931

= 28.28 MeV

∴ Binding energy per nucleon = \(\frac{28.28}{4}=7.07\,MeV\)

Answer 2

Mass defect=[2.01565+2.01733-4.002604]
=0.030376 amu
Binding energy =(0.030376 x 931 ) MeV
=28.28 MeV
`therefore` Binding energy per nucleon =`28.28/2`=7.07 MeV