An electron is trapped in a quantum dot, in which it is continued to a very small region in all three dimensions, If the lowest energy transition is to produce a photon of $\mathbf{450}\mathbf{\text{nm}}$ wavelength, what should be the width of the well (assumed cubic)?

The transition is between the group state with quantum numbers $\left(n_{\text{x}},n_{\text{y}},n_{\text{z}}\right)\mathbf{=}\left(1,1,1\right)$and to one of three states of the next highest energy with quantum numbers.

$\left(n_{\text{x}},n_{\text{y}},n_{\text{z}}\right)\mathbf{=}\left(2,1,1\right)\mathbf{,}\left(1,2,1\right)\mathbf{\text{or}}\left(1,1,2\right)$

Write the expression for photon using the following relation.

$E=\frac{hc}{\lambda }$ ….. (1)

Here,

The photon’s wavelength, $\lambda =450\text{nm}=450\times {10}^{-9}\text{m}$

Planck’s constant, $h=6.62607\times {10}^{-34}\text{J}·\text{s}$

The speed of light, $c=2.9979\times {10}^8\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$

Substitute known values into equation (1).

$\begin{array}{rcl}E& =& \frac{\left(6.62607\times {10}^{-34}\text{J}·\text{s}\right)\left(2.9979\times {10}^8{\displaystyle \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.}\right)}{450\times {10}^{-9}\text{m}}\\ & =& 4.4143\times {10}^{-19}\text{J}\end{array}$

The above result is equal the difference of the first excited state energy and grounded state energy which is given by,

$\begin{array}{rcl}{E}_{1,1,2}-E_{1,1}& =& \left(1^2+1^2+2^2\right)\frac{{\pi }^2{h}^2}{2m{L}^2}-\left(1^2+1^2+1^2\right)\frac{{\pi }^2{h}^2}{2m{L}^2}\\ & =& 3\frac{{\pi }^2{h}^2}{2m{L}^2}\end{array}$

Now, set the transition energy equal to the energy of photon.

$$\begin{gathered} E_{1,1,2}-E_{1,1}=\frac{h\lambda }{c}\backslash \mathrm{hfill} \\ 3\frac{{\pi }^2{h}^2}{2m{L}^2}=4.4143\times {10}^{-19}\text{J} \\ L=\sqrt{3\frac{{\pi }^2{h}^2}{2m}\frac{1}{4.4143\times {10}^{-19}\text{J}}} \end{gathered}$$

Substitute $9.1094\times {10}^{-31}\text{kg}$for the mass of the electron $m$, $1.054\times {10}^{-34}\text{J}·\text{s}$ for $h$in above equation.

$\begin{array}{rcl}L& =& \sqrt{3\frac{{\left(3.14\right)}^2{\left(1.054\times {10}^{-34}\text{J}·\text{s}\right)}^2}{2\left(9.1094\times {10}^{-31}\text{kg}\right)}\frac{1}{4.4143\times {10}^{-19}\text{J}}}\\ & =& 6.40\times {10}^{-10}\text{m}\\ & =& \left(6.40\times {10}^{-10}\text{m}\right)\left(\frac{\text{nm}}{{10}^{-9}\text{m}}\right)\\ & =& 0.640\text{nm}\end{array}$

Hence, the width should be $0.640\text{nm}$.