Q21E - University Physics with Modern Physics

Question

Question: 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)?

Step-by-Step Solution

Answer

The width should be  $0.640 \text{nm}$.

Step 1: Concept

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

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

Step 2: Determine photon energy:

Write the expression for the 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).

$\begin{array}{rcl}E& =& \frac{\left(6.62607\times {10}^{-34} \text{J}·\text{s}\right)\left(2.9979\times {10}^8 \raisebox{1ex}{$\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 tthe photon.

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

Substitute $9.1094\times {10}^{-31} \text{kg}$ for the mass of the electron $m$ and $1.054\times {10}^{-34} \text{J}·\text{s}$ for $h$ in the 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 $0.640 \text{nm}$.