Suppose you put five electrons into a $0.50nm$-wide one dimensional rigid box (i.e., an infinite potential well).

a. Use an energy-level diagram to show the electron configuration of the ground state.

b. What is the ground-state energy that is, the total energy of all five electrons in the ground-state configuration?

$E_1=\frac{h^2}{8m{L}^2}$

Here, $h$is the planck's constant. $m$is the mass of the electron, and $L$is the length of the rigid box.

Convert the units for the length of the rigid box from $nm$to $m$.

$$\begin{gathered} L=0.50nm \\ =0.50nm\left(\frac{1m}{1\times {10}^{9}nm}\right) \\ =0.50\times {10}^{-9}m \end{gathered}$$

Substitute $0.5\times {10}^{-9}m$for $L,6.63\times {10}^{-34}J.s$for $h,9.11\times {10}^{-31}kg$for $m_c$in equation

$E_1=\frac{h^2}{8m{L}^2}$to obtain the ground state energy of the electron in the box.

$$\begin{gathered} E_1=\frac{{\left(6.63\times {10}^{-34}J.s\right)}^2}{8\left(9.11\times {10}^{-31}kg\right){\left(0.5\times {10}^{-9}m\right)}^2} \\ =2.41\times {10}^{-19}J \end{gathered}$$

Convert the energy from $J$to $eV$.

$$\begin{gathered} E_1=2.41\times {10}^{-19}J\left(\frac{1eV}{1.6\times {10}^{-19}J}\right) \\ =1.51eV \end{gathered}$$

The ground state energy of the$0.5nm$wide$1D$rigid boxes is$1.51eV$.

The higher energy state with quantum state $n$is

Only two electrons can be placed in one energy state. The next energy state has quantum state $n=2$. Substitute $E_1=1.51eV$and $n=2$to obtain the next energy state:

$$\begin{gathered} E_2=2^2(1.51eV) \\ =6.04eV \end{gathered}$$

Now we need one more energy state to place the last electron. Substitute $E_1=1.51eV$and $n=3$to obtain the next energy state:

$$\begin{gathered} E_3=3^2(1.51eV) \\ =13.6eV \end{gathered}$$

The energy level diagram of the configuration is shown below:

Calculate the ground state energy of the one dimensional rigid box.

The ground state energy of the five electrons is,

$E_{ground=2E_1+2E_2+E_3}$

Substitute $1.51eV$for $E_1,6.04eV$for $E_2$and $13.6eV$for $E_3$.

$$\begin{gathered} E_{ground}=2E_1+2E_2+E_3 \\ =2(1.51eV)+2(6.04eV)+13.6eV \\ =28.7eV \end{gathered}$$

Therefore, the ground-state energy of the configuration is$28.7eV$