Question

Figure 40-23 is an energy-level diagram for a fictitious infinite potential well that contains one electron. The number of degenerate states of the levels is indicated: “non” means non-degenerate (which includes the ground state of the electron), “double” means 2 states, and “triple” means 3 states. We put a total of 11 electrons in the well. If the electrostatic forces between the electrons can be neglected, what multiple of ${\mathbf{h}}^{\mathbf{2}}\mathbf{/}\mathbf{8}{\mathbf{mL}}^{\mathbf{2}}$gives the energy of the first excited state of the 11 electron system?

Short answer

The multiple of${\mathrm{h}}^2/8{\mathrm{mL}}^2$ that gives the energy of the first excited state of the 11 electron system is 66.

Step-by-step solution

The given data:

A total of 11 electrons are put in the fictitious potential well system that contains one electron and levels like non-degenerate, 2 states and three states.

Understanding the concept of spin degeneracy

Pauli's exclusive principle states that no two electrons in the same atom can have the same values in all four of their quantum numbers.

"Spin deterioration" here means that each energy level is able to capture the electron spin-up and the electron spin-down, i.e. in each orbital there are two distinct rotating states.

Considering the concept of electron accommodation and spin degeneracy that is Pauli's exclusion principle, that use to define the number of electrons in each energy level at the ground state and the first excited state. Thus, the ground state has two electrons in the non-degenerate states, 6 in the triplet state, and 3 in the double state to get the lowest possible energy. Similarly, the first excited state has two electrons in the non-degenerate states, 5 in the triplet state, and 4 in the double state.

Calculation of the multiple of h2/8mL2 for the first excited state

Using the concept of spin degeneracy $\left({\mathrm{m}}_{\mathrm{s}}=\pm \frac{1}{2}\right)$, each state can accommodate only two electrons, one of spin-up and the other of spin-down.

Now, from the given diagram, the energy of each state is given as below.

The energy of ground state is

${\mathrm{E}}_{ground}\left({\mathrm{E}}_1\right)=\frac{4{\mathrm{h}}^2}{8{\mathrm{mL}}^2}$

The energy of double state is,

${\mathrm{E}}_{double}\left({\mathrm{E}}_2\right)=\frac{6{\mathrm{h}}^2}{8{\mathrm{mL}}^2}$

The energy of triple state is,

${\mathrm{E}}_{\mathrm{triple}}\left({\mathrm{E}}_3\right)=\frac{7{\mathrm{h}}^2}{8{\mathrm{mL}}^2}$

Now, for the 11 electron system in the ground state, using the concept the total energy can be given as follows:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{excited},\mathrm{total}}=2{\mathrm{E}}_1+6{\mathrm{E}}_2+3{\mathrm{E}}_3 \\ =2\left(\frac{4{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right)+6\left(\frac{6{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right)+3\left(\frac{7{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right) \\ =65\left(\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right) \end{gathered}$$

But, the same 11 electron system in the first excited state can given a total minimum possible energy as follows:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{excited},\mathrm{total}}=2{\mathrm{E}}_1+5{\mathrm{E}}_2+4{\mathrm{E}}_3 \\ =2\left(\frac{4{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right)+5\left(\frac{6{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right)+4\left(\frac{7{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right) \\ =66\left(\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right) \end{gathered}$$

Hence, the value of the multiple is 66.