Question

Calculate the binding energy E of the helium nucleus 42He (1eV=1.602×10−19J). Express your answer in millions of electron volts to four significant figures.

Answer 1

From the given atom  ₂He⁴  ,We Could see that No of protons= 2 , No of nucleons= 4.

So ,no of neutrons= 4 -2 = 2 .

Actual Atomic mass of helium = 4 amu= 6.6447 x 10^-27 kg.

First we find individual mass of nucleons in this nucleus

= mass of 2 protons+ mass of 2 neutrons

= 2 (mass of proton )+ 2( mass of neutron)

= 2( 1.6726x 10^{- 27} kg ) + 2 ( 1.6749 x 10^{- 27} kg)

= 6.695 x 10^-27 Kg .

Mass defect of helium nucleus= Calculated atomic mass - Actual atomic mass

M= (6.695 - 6.6447 ) x 10^-27 kg

= 0.0503 x 10^-27 kg.

This mass defect is used as binding energy to bind nucleons tightly in nucleus.

Binding energy ( E) = MC²

( C= Velocity of light in vaccum. )

E= 0.0503 x 10^-27 x (3 x 10⁸ )²

= 4.527 x 10 ^{- 12} J

= (4.527 x 10^-12 )/ 1.602 x 10^-19 eV.

= 28.26x 10⁶ eV

= 28.26 MeV .( million electron Volts)