Question

The first two terms in the semi empirical binding energy formula deal solely with the internucleon attraction.

(a) Calculate the ratio of the second term to first term for$A=20$ . What does it say about the surface nucleons?

(b) Repeat part (a), but for$A=220$ .

Short answer

(a) The resultant answeris $-0.42$.

(b) The resultant answeris $-0.19$.

Step-by-step solution

Given data

$A=20$And $A=220$is the given data.

Formula of Semi empirical binding Energy

Semi empirical binding Energy is given by the expression,$BE=c_{1}A-c_2{A}^{2/3}-c_3\frac{Z(Z-1)}{A^{1/3}}-c_4\frac{{(N-Z)}^2}{A}$ .

Where,$c_1=15.8MeV,c_2=17.8MeV,c_3=0.71MeV,\text{\hspace{0.17em}}{c}_4=23.7MeV$.

$A\to$Mass Number,$Z\to$atomic Number and$N\to$number of neutrons.

Calculate the semi empirical binding energy

(a)

The ratio of the second term to the first term in semi empirical binding Energy formula is:

$\begin{array}{c}\frac{\text{Secondterm}}{\text{Firstterm}}=-\frac{c_2{A}^{2/3}}{c_{1}A}\\ \frac{\text{Secondterm}}{\text{Firstterm}}=-\frac{17.8}{15.8A^{1/3}}\\ \frac{\text{Secondterm}}{\text{Firstterm}}=-\frac{17.8}{15.8{\left(20\right)}^{1/3}}\\ \frac{\text{Secondterm}}{\text{Firstterm}}=-0.42\end{array}$

The surface term is a fraction of the volume term and it is negative.

The surface nucleons have fewer nucleons surrounding them and attracting them

Calculate the semi empirical binding energy

(b)

If $A=220$ the ratio is:

$\begin{array}{l}\frac{\text{Secondterm}}{\text{Firstterm}}=-\frac{c_2{A}^{2/3}}{c_{1}A}\\ \frac{\text{Secondterm}}{\text{Firstterm}}=-\frac{17.8}{15.8A^{1/3}}\\ \frac{\text{Secondterm}}{\text{Firstterm}}=-\frac{17.8}{15.8{\left(220\right)}^{1/3}}\\ \frac{\text{Secondterm}}{\text{Firstterm}}=-0.19\end{array}$

The surface term is a fraction of the volume term and it is negative.

The surface nucleons have fewer nucleons surrounding them and attracting them.