Question

An electron in the n=4 state of a 5 nm wide infinite well makes a transition to the ground state, giving off energy in the form of photon. What is the photon’s wavelength?

Short answer

The wavelength of the photon is $5.5\times {10}^{-6}$.

Step-by-step solution

Writing the energy transition equation.

We know that energy for an infinite well is given by

${\mathrm{E}}_{\mathrm{n}}=\frac{{\mathrm{n}}^2{\mathrm{\pi h}}^2}{2{\mathrm{mL}}^2}$

Since the electron is making transition from n = 4 to n = 1 state

$$\begin{gathered} {\mathrm{E}}_4{\mathrm{E}}_1=\frac{{\mathrm{\pi }}^2{\mathrm{h}}^2}{2\mathrm{mL}}\left(4^2-1^2\right) \\ =\frac{15{\mathrm{\pi }}^2{\mathrm{h}}^2}{2\mathrm{mL}} \end{gathered}$$

Relating energy with wavelength.

We have $\begin{array}{l}\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}\\ \mathrm{\lambda }=\frac{\mathrm{hc}}{\mathrm{E}}\end{array}$

Substituting E from above equation

$\mathrm{\lambda }=\frac{2{\mathrm{mL}}^2\mathrm{hc}}{15{\mathrm{\pi }}^2{\sout{\mathrm{h}}}^2}$

Arriving to the photon wavelength.

Substituting all the values, $\mathrm{L}=5\mathrm{nm},\mathrm{m}=9.1\times {10}^{-31}\mathrm{kg}$

$$\begin{gathered} \mathrm{\lambda }=\frac{2\left(9.1\times {10}^{-31}\right){\left(5\times {10}^{-9}\right)}^2\left(6.625\times {10}^{-34}\right)\left(3\times {10}^8\right)}{15\mathrm{\pi }\left(1.05\times {10}^{-34}\right)} \\ =5.5\times {10}^{-6}\mathrm{m} \end{gathered}$$