Question

The mass of helium nucleus is less than that of its constituent particle by 0.03 a.m.u find binding energy per nucleon

Answer 1

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Answer 2

Answer:

The binding energy per nucleon of Helium is equal to 7MeV.

Explanation:

• Mass defect:- The sum of the mass of neutrons and protons forming a nucleus is more than the actual mass of the nucleus. The difference of predicted mass and actual mass of the nucleus is called mass defect.
• Binding energy :- The energy required to combine the neutrons and protons into a nucleus is called binding energy.

We have the mass defect of helium nucleus = 0.03 a.m.u

Δm = 0.03×1.67×10⁻²⁷kg

Δm = 0.0501×10⁻²⁷kg

Binding energy is given by:

E = Δmc²     where c is velocity of light

E= (0.0501×10⁻²⁷kg)×(3×10⁸ms⁻¹)²

E = 0.4509×10⁻²⁷⁺¹⁶ kgm²s⁻²

E = 0.4509×10⁻¹¹J

$E=\frac{0.4509*10^{-11}}{1.6*10^{-19}}eV$

$E=0.2818*10^{8}eV$

$E=28.18MeV$

The binding energy per nucleon is given by:-

Binding energy per nucleon $=\frac{Binding\;\; energy}{number \; \; of\; \; nucleons}$

Number of nucleons of He = 4

Binding energy per nucleon $=\frac{28.18}{4} = 7MeV$