91影视

A total of 22 electrons are put in the fictitious potential well system that contains one electron and levels like non-degenerate, states, and three states.

Spin degradation "here means that each level of energy is able to capture the electron spin-up and the electron spin-down, i.e. in each orbital there are two distinct spinning regions (some may call these spins orbitals).

Considering the concept of electron accommodation and spin degeneracy which is Pauli's exclusion principle, we can get the number of electrons in each energy level at the ground state. Due to spin degeneracy, each state can have two electrons. For the case being 2 electrons in the ground state, 6 electrons in the triplet state, and 2 electrons in the non-degenerate states, we can get the total energy of the system.

Thus using the concept of spin degeneracy $\left(m_s=卤\frac{1}{2}\right)$for a 22 electron system, the number of electrons in the ground is 2, the triple-state is 6, the higher triple-state is 6, another 6 in the next higher triple-state, and finally the last two electrons in the no-degenerate states. From figure, the energy values of each state are given as:

$E_{ground}\left(E_1\right)=\frac{3h^2}{8m{L}^2}$

$$\begin{gathered} E_{triple}\left(E_2\right)=\frac{6h^2}{8m{L}^2} \\ {E}_{triple}\left(E_3\right)=\frac{9h^2}{8m{L}^2} \\ {E}_{triple}\left(E_4\right)=\frac{11h^2}{8m{L}^2} \\ {E}_{non-degenerate}\left(E_5\right)=\frac{12h^2}{8m{L}^2} \end{gathered}$$

Thus, the total energy of the 22 electron system for the ground state is given as follows.

localid="1661505572735" $$\begin{gathered} E_{ground,total}=2E_1+6E_2+6E_3+6E_4+2E_5 \\ =2\left(\frac{3h^2}{8m{L}^2}\right)+6\left(\frac{6h^2}{8m{L}^2}\right)+6\left(\frac{9h^2}{8m{L}^2}\right)+6\left(\frac{11h^2}{8m{L}^2}\right)+2\left(\frac{12h^2}{8m{L}^2}\right) \\ =186\left(\frac{h^2}{8m{L}^2}\right) \end{gathered}$$

Hence, the value of the multiple is 186.