Three electrons are in a one-dimensional rigid box (i.e., an infinite potential well) of length $0.50nm$. Two are in the $n=1$state and one is in the localid="1649073297418" $n=6$state. The selection rule for the rigid box allows only those transitions for which $∆n$is odd.

a. Draw an energy-level diagram. On it, show the filled levels and show all transitions that could emit a photon.

b. What are all the possible wavelengths that could be emitted by this system?

$$\begin{gathered} E_{n=0}=1.51{\left(0\right)}^{2}eV=0 \\ {E}_{n=1}=1.51{\left(1\right)}^{2}eV=1.51eV \\ {E}_{n=2}=1.51{\left(2\right)}^{2}eV=6.04eV \\ {E}_{n=3}=1.51{\left(3\right)}^{2}eV=13.59eV \\ {E}_{n=4}=1.51{\left(4\right)}^{2}eV=24.16eV \\ {E}_{n=5}=1.51{\left(5\right)}^{2}eV=37.75eV \\ {E}_{n=6}=1.51{\left(6\right)}^{2}eV=54.36eV \end{gathered}$$

Now, the allowed transitions are as follows: $$\begin{gathered} E_6\to E_5,E_6\to E_3,E_6\to E_1 \\ {E}_5\to E_4,E_5\to E_2 \\ {E}_4\to E_3,E_4\to E_1 \\ {E}_3\to E_2 \\ {E}_2\to E_1 \end{gathered}$$

Wavelength of the emitted photon is,

$\lambda =\frac{1240eVnm}{∆Eev}$

Here, $\lambda$is the wavelength.

For transition:

$E_6\to E_5,\lambda =\frac{1240eVnm}{54.36eV-37.75eV}=74.65nm$

For transition:

$E_6\to E_3,\lambda =\frac{1240eVnm}{54.36eV-13.59eV}=30.41nm$

For transition:

$E_6\to E_1,\lambda =\frac{1240eVnm}{54.36eV-1.51eV}=23.46nm$

For transition:

$E_5\to E_4,\lambda =\frac{1240eVnm}{37.75eV-24.16eV}=391.24nm$

For transition

$E_5\to E_2,\lambda =\frac{1240eVnm}{37.75eV-6..04eV}=39.1nm$

For transition

$E_4\to E_3,\lambda =\frac{1240eVnm}{24.16eV-13.59eV}=117.3nm$

For transition

$E_4\to E_1,\lambda =\frac{1240eVnm}{24.16eV-1.51eV}=54.75nm$

For transition

$E_3\to E_2,\lambda =\frac{1240eVnm}{13.59eV-16.04eV}=164.24nm$

For transition

$E_2\to E_1,\lambda =\frac{1240eVnm}{6.04eV-1.51eV}=273.73nm$

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

Length of the box is,

$$\begin{gathered} L=0.5nm \\ =(0.5nm)({10}^{-9}m/nm) \\ =0.5\times {10}^{-9}m \end{gathered}$$

$$\begin{gathered} E_n=n^2\left(\frac{{\left(6.63\times {10}^{-34}Js\right)}^2}{\left(8\right)\left(9.1\times {10}^{-31}kg\right){\left(0.5\times {10}^{-9}m\right)}^2}\right) \\ =\left(2.415\times {10}^{-19}J\right)n^2 \\ =\frac{\left(2.415\times {10}^{-19}J\right)n^2}{\left(1.6\times {10}^{-19}{\displaystyle \frac{J}{eV}}\right)} \\ =1.51n^{2}eV \end{gathered}$$