Seven electrons are trapped in a one-dimensional infinite potential well of width L. What multiple of ${\mathbf{h}}^{\mathbf{2}}\mathbf{/}\mathbf{8}{\mathbf{mL}}^{\mathbf{2}}$ gives the energy of the ground state of this system? Assume that the electrons do not interact with one another, and do not neglect spin.

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

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 that further give the multiple of ${\mathbf{h}}^{\mathbf{2}}\mathbf{/}\mathbf{8}{\mathbf{mL}}^{\mathbf{2}}$.

Formula:

The energy of ${\mathrm{n}}^{\mathrm{th}}$state of a ground state of the system is,

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

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

To get the lowest possible total energy, two electrons are filled in each state of $n=1,2,3$, while the rest of one electron fills the state of n = 4.

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

role="math" localid="1661497763518" $$\begin{gathered} {\mathrm{E}}_{\mathrm{ground}}=2{\mathrm{E}}_1+2{\mathrm{E}}_2+2{\mathrm{E}}_3+{\mathrm{E}}_4 \\ =2\left(\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right)+2\left(\frac{2^2{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right)+2\left(\frac{3^2{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right)+2\left(\frac{4^2{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right) \\ =\left(2+8+18+16\right)\left(\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right) \\ =\left(44\right)\left(\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right) \end{gathered}$$

Hence, the value of the multiple is 44.