For the situation of Problem 21, what multiple of ${\mathbf{h}}^{\mathbf{2}}\mathbf{/}\mathbf{8}{\mathbf{mL}}^{\mathbf{2}}$gives the energy of (a) the first excited state, (b) the second excited state, and (c) the third excited state of the system of seven electrons? (d) Construct an energy-level diagram for the lowest four energy levels of the system.

There are seven electrons trapped in a one-dimensional infinite potential well of width L.

The low state of the quantum-mechanical system is its fixed state of extremely low power; the power of the lower state is known as the zero power of the system.

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.

Using the concept of Pauli's exclusion principle of an infinite potential well, distribute the electrons accordingly to get the smallest value of energy in the ground state of the system. Thus, the required total energy due to all these electrons further gives the multiple of. Using similar values and the condition of the excited state, the electrons occupying each state; thus, the total energy in each excited state.

Formula:

The energy of nth state of a ground state of the system is,

$E_n=\frac{{\mathrm{n}}^2{\mathrm{h}}^2}{8\mathrm{mL}}$ ….. (1)

Here, n is the number of states, h is the Plank’s constant, m is the mass, and L is the width.

To get the next not-fully occupied higher level that is the first excited state, the configuration for the least total energy greater than that of the ground state has the two electrons are filled in each state of n=1 ,2 the third state n=3 half-filled while the state of n=4 has the remaining two electrons.

Now using the n values in energy equation (1), the total energy of the ground state of the system is given as follow.

$$\begin{gathered} {\mathrm{E}}_{\mathrm{first}\mathrm{excited}}=2{\mathrm{E}}_1+2{\mathrm{E}}_2+{\mathrm{E}}_3+2{\mathrm{E}}_4 \\ =2\left(\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right)+2\left(\frac{2^2{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right)+\left(\frac{3^2{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right)+2\left(\frac{4^2{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right) \\ =\left(2+8+9+32\right)\left(\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right) \\ =51\left(\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right) \end{gathered}$$

Hence, the value of the multiple is 51.

To get the next not-fully occupied higher level that is the second excited state, the configuration for the least total energy has the two electrons are filled in each state of n=1,2,3, while the remaining seventh electron is in the state of n=5.

Now using the values in energy equation (1), the total energy of the ground state of the system is given as follow.

$$\begin{gathered} {\mathrm{E}}_{second\mathrm{excited}}=2{\mathrm{E}}_1+2{\mathrm{E}}_2+{\mathrm{E}}_3+2{\mathrm{E}}_4 \\ =2\left(\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right)+2\left(\frac{2^2{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right)+\left(\frac{3^2{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right)+2\left(\frac{4^2{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right) \\ =\left(2+8+18+25\right)\left(\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right) \\ =53\left(\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right) \end{gathered}$$

Hence, the value of the multiple is 53.

To get the next not-fully occupied higher level that is the third excited state, the configuration for the least total energy has the two electrons are filled in each state of n=1,2,3, while the state of n=2 being half-filled.

Now using the values in energy equation (1), the total energy of the ground state of the system is given as follow.

$$\begin{gathered} {\mathrm{E}}_{third\mathrm{excited}}=2{\mathrm{E}}_1+2{\mathrm{E}}_2+{\mathrm{E}}_3+2{\mathrm{E}}_4 \\ =2\left(\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right)+2\left(\frac{2^2{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right)+\left(\frac{3^2{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right)+2\left(\frac{4^2{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right) \\ =\left(2+8+18+32\right)\left(\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right) \\ =56\left(\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right) \end{gathered}$$

Hence, the value of the multiple is 56 .

The energy states of this problem are given below in the diagram as: