Mirror Nuclei and Coefficient of the Surface Energy

Those nuclei in which the number of protons in one is equal t the number of neutrons in another are called mirror-nuclei. The mass number of the mirror nuclei are same and the number of protons differs by one unit.

For example $$_1H^3 and _2He^3$$

Let 'R' be the common radius off two neighboring mirror nuclei N₁ and N₂. Since protons are uniformly distributedin the entire volume of the nucleus, the coulombs potential energy of the z-protons in nucleus N₁is given by;

$$(E_c)N_1 = \frac 35 \frac {z^2 e^2}{R} (in C.G.S)$$

Since N_z nucleus contains one more protons i.e. Z+1, the coulomb potential energy of (Z+1) protons is given by,

$$(E_c)N_2 =\frac 35 \frac {(z+1)^2 e^2}{R} (in C.G.S)$$

Thus, the difference between the coulomb potential energies of two neighboring mirror nuclei is given by;

$$\Delta E_c =(E_c)N_2-(E_c)N_1$$

$$=\frac 35 \frac{e^2}{R}(2Z+1)$$

Since 2Z+1=A i.e mass number of mirror nuclei.

$$\Delta E_c =\frac 35 \frac{e^2}{R} A$$

Also,

$$\Delta E_c =\frac 35 \frac{e^2}{R_0}A^{1-1/3};R=R_0A^1/3$$

$$\Delta E_c =\frac 35 \frac{e^2}{R_0}A^{2/3}$$

Which is the difference between coulomb energies of two neighbouring mirror nuclei.

A rough estimate of the coefficients of the surface energy term in the theory of semi-empirical mass formula.

The molecules on the surface of the liquid are less bound than the inner molecules and these have the tendencyto go out. so the surface molecules have a tendency to decrease the total binding energy. We expect to the tendency of decrease of the total binding energy of molecules due to surface nucleons and is proportional to the surfaces area of the nucleus.

$$-B_s \alpha 4\pi R^2=4\pi R_0^2 A^{2/3}$$

$$\therefore B_s = -a_s A^{2/3}...........................(i) $$

Where a, is constant and it is called the surface constant. \(a_s\) =13 MeV, experimentally

We know

Volume energy \(B_v =a_v A\) ............................. (ii)

\(A_v\) is the volume constant.

The space available to a single nucleon in the nucleons is \(\frac 43 \frac{\pi R^3}{A}\)

Now, we imagine a nuclear surface region of thickness 't'

$$t=\sqrt [3] {\frac 43}\frac{\pi R^3{A}}...........................(3)$$

Then the surface region must contain approximately,

$${\frac{\frac 43}{\pi R^3A}}$$