Question

An electron is trapped in a quantum well (practically infinite). If the lowest-energy transition is to produce a photon ofwavelength. Whatshould be the well’s width?

Short answer

The width of the well is $6.4\times {10}^{-10}\text{m}$.

Step-by-step solution

Formula used.                  

The lowest transition of the electron produces a photon of wavelength.

Formula used:

The expression for the difference in energy levels is given by

$\mathbf{\Delta E}\mathbf{=}{\mathbf{E}}_{\mathbf{2}}\mathbf{-}{\mathbf{E}}_{\mathbf{1}\mathbf{}}$

Here,$E_2$and$E_1$are the energy of the highest and lowest energy levels, respectively.

The expression for the energy of thenthlevel is given by

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

Here,nis the number of energy levels.,Is reduced Planck’s constant,$\hslash =h^2\pi$.

h is Planck’s constant,$h=6.64\times {10}^{-34}\text{J}·\text{s}$,m is the mass of an electron$9.1\times {10}^{-31}\text{kg}$, andLis the width of the quantum well.

Determine the equation in terms of the width of the quantam well as:

$\begin{array}{rcl}\frac{hc}{\lambda }& =& \frac{2^2{\pi }^2{h}^2}{2m{L}^2}-\frac{1^2{\pi }^2}{2m{L}^2}\\ \frac{hc}{\lambda }& =& \frac{3{\pi }^2{h}^2}{2m{L}^2}\\ L& =& \sqrt{\frac{3{\pi }^{2}h\lambda }{8{\pi }^{2}mc}}\\ & & \end{array}$

Determine the width of the well            

Substitute the values and solve as:

$\begin{array}{rcl}L& =& \sqrt{\frac{3{\pi }^{2}h\lambda }{8{\pi }^{2}mc}}\\ & =& \sqrt{\frac{3{\pi }^2\left(6.625\times {10}^{-34}\right)\left(450\times \frac{1}{{10}^9}\right)}{8{\pi }^2\left(9.1\times {10}^{-31}\right)\left(3.0\times {10}^8\right)}}\\ & =& 0.64\text{nm}\\ & & \end{array}$

The width of the quantum well in which an electron is trapped, which produces a photon of wavelength for its lowest transition is$6.4\times {10}^{-10}\text{m}$