Question

In heavy nuclei, number of neutrons is greater than the number of protons .WHY​

Answer 1

Answer:

Explanation:

Very roughly speaking, it’s because the Coulomb energy of a nucleus is repulsive and goes up as  Z2A1/3  ,  where  A=Z+N,   Z  is the number of protons and  N  is the number of neutrons, while the attractive binding energy of a nucleus goes up as  A.  The strong nuclear force is attractive at intermediate to long range, it turns strongly repulsive at very short range, and it also has a finite range, so in a finite nucleus the attractive force eventually saturates, growing as the volume or mass number of a nucleus, while the repulsive Coulomb force has infinite range and it grows with the square of the number of protons and drops only slowly with the radius of the nucleus. So in a large nucleus with many protons, the Coulomb force will win over the strong nuclear force, unless the number of neutrons  N  increases more rapidly than the number of protons  Z.  

To a first approximation you can write an expression for the binding energy of a nucleus, relative to free protons and neutrons, as:

B(Z,N)=aVA−aCZ(Z−1)A1/3.  

It’s more complicated but this is the main idea. There is also an asymmetry energy that comes due to the exclusion principle. But in the absence of the Coulomb repulsion, and any neutron proton mass difference, the strong nuclear forces would favor equal numbers of protons and neutrons.

The maximum binding energy shifts to smaller  Z/N  as  A  increases, and the more stable nuclei are found near to the maximum.