Question

If the total number of nucleons in a nucleus is 'A', then the surface effect reduces its binding energy, by a factor directly proportional to:
1. \({A^{\frac{1}{3}}}\)
2. \({A^{\frac{1}{2}}}\)
3. \(A^2\)
4. \({A^{\frac{2}{3}}}\)

Answer 1

Correct Answer - Option 4 : \({A^{\frac{2}{3}}}\)

Concept:

Liquid Drop Model of Nucleus

• A nucleus is considered to be similar to a liquid drop due to similarity in nature. 
• Due to surface tension, the liquid drop tends to be spherical. This is because the 
• A liquid drop contains liquid molecules and a nucleus contains nucleons in the very same manner. 
• The surface of the liquid has lesser energy and has more tendency to leave the drop. Similarly, the nucleons at the surface have more tendency to leave the nucleus. 
• Practically this effect is called the surface effect. 

Relationship between Number of Nucleons and Surface Energy

The radius of the nucleus is proportional to the one-third power of a number of nucleons A.

R = R₀ A ^1/3

Explanation:

The surface effect is related to the surface area of the nucleus. 

More the weaker binding energy of the nucleus larger the surface area and the surface energy. 

The surface Area of the sphere is E = 4 π R²

But, R = R₀ A ^1/3

So, we can say Surface area = E = 4 π ( R₀ A 1/3) 2 = K A ^2/3

where K is a constant

Or, E ∝ A ^2/3

So as surface energy increases, the binding energy decreaes by same ratio. 

So, we can say that it has been reduced by factor of \({A^{\frac{2}{3}}}\)