Question

The mass defect for the nucleus of helium is 0.0303 a.M.U. What is the binding energy per nucleon for helium in mev?

Answer 1

Answer:

According to Einstein's Special Theory of Relativity, Energy equivalent to this Mass ΔM is E = ΔM × c²

⇒ But as ΔM is the Mass Defect, The Energy equivalent to ΔM is the Energy required to break the respective nucleus into Protons and Neutrons. (i.e.) The Energy equivalent to ΔM is the Binding energy of the Respective Nucleus.

⇒ E(b) = ΔM × c²

We know that 1 Atomic mass unit (u) = 1.66 × 10⁻²⁷ kg

Let us find Energy equivalent to 1 Atomic mass unit (u) :

E = mc²

E(1u) = 1.66 × 10⁻²⁷ × [3 × 10⁸]²

E(1u) = 1.49 × 10⁻¹⁰ J

We know that 1eV = 1.602 × 10⁻¹⁹ J

Converting E(1u) in Joules to Electronvolts :

E(1u) = (1.49 × 10⁻¹⁰)/(1.6 × 10⁻¹⁹) = 0.9315 × 10⁹ eV = 931.5 MeV

⇒ Energy equivalent to 1 Atomic mass unit (u) = 931.5 MeV (This should be remembered as a standard result)

Given that the Mass Defect of Helium Nucleus is 0.0303 amu

⇒ Binding Energy of Helium Nucleus = 0.0303 × 931.5 MeV

⇒ Binding Energy of Helium Nucleus = 28.2244 MeV

But this is the Binding Energy of Helium Nucleus, We know that the Helium Nucleus contains 2 Protons and 2 Neutrons (i.e.) ⇒ 4 Nucleons.

⇒ Binding Energy per Nucleon = (28.2244)/4 = 7.056 MeV