Question

What is meant by binding energy? If the masses of proton, neutron and alpha $\left(\alpha \right)$ particles are $1.00728amu,1.00867amu$ and $4.00150amu$ respectively, then find the binding energy per nucleon of alpha particle. $[1amu=931MeV]$.

Answer 1

Binding energy of nucleus is defined as the energy that holds the components of nucleus together or is also defined as the amount of energy required to break the nucleus completely into its constituent particles.
Mathematically, binding energy is equal to the product of mass defect of nucleus and the square of speed of light.
Binding energy $E_B=\Delta M\times c^2=\Delta M\left(inamu\right)\times 931MeV$
Alpha particle has 2 protons and 2 neutrons. So alpha particle has 4 nucleons.
Mass defect of alpha particle $\Delta M=2m_p+2m_n-m_{He}=2\left(1.00728\right)+2\left(1.00867\right)-4.00150=0.0304MeV$
Thus binding energy $E_B=\Delta M\times 931=0.0304\times 931=28.3MeV$
Binding energy per nucleon $=\frac{E_B}{4}=\frac{28.3}{4}=7.0756MeV$