An electron in a long, organic molecule used in a dye laser behaves approximately like a particle in a box with width 4.18 nm. What is the wavelength of the photon emitted when the electron undergoes a transition (a) from the first

excited level to the ground level and (b) from the second excited level to the first excited level?

The energy of a particle in a box is,

$$\begin{gathered} \mathrm{E}=\frac{{\mathrm{n}}^2{\mathrm{TT}}^2{\mathrm{h}}^{-2}}{2{\mathrm{mL}}^2} \\ =\frac{{\mathrm{n}}^2{\mathrm{h}}^2}{8{\mathrm{mL}}^2} \end{gathered}$$ ...(i)

Also, the energy-wavelength relation is,

$∆\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}$ ...(ii)

Now, for first exited level n = 2 and ground level n = 1, the energy is,

${\mathrm{E}}_2=\frac{4{\mathrm{h}}^2}{8{\mathrm{mL}}^2}$and${\mathrm{E}}_1=\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}$

So,

$$\begin{gathered} ∆\mathrm{E}={\mathrm{E}}_2-{\mathrm{E}}_1 \\ =\frac{3{\mathrm{h}}^2}{8{\mathrm{mL}}^2} \end{gathered}$$

Substitute the value of $∆\mathrm{E}$in equation (ii) and solve for the wavelength,

$$\begin{gathered} \frac{3{\mathrm{h}}^2}{8{\mathrm{mL}}^2}=\frac{\mathrm{hc}}{\mathrm{\lambda }} \\ =\frac{8{\mathrm{mcL}}^2}{3\mathrm{h}} \\ =\frac{8\left(9.109\times {10}^{-3}\mathrm{kg}\right)\left(2.998\times {10}^8\mathrm{m}/\mathrm{s}\right){\left(4.18\times {10}^{-9}\mathrm{m}\right)}^2}{3\left(6.626\times {10}^{-34}\mathrm{J}-\mathrm{s}\right)} \\ =1.92\times {10}^{-5}\mathrm{m} \\ =19.2\mathrm{\mu m} \end{gathered}$$

Thus, the wavelength from this transition is 19.2 µm.

Similarly, for first exited level n = 2 and second excited level n = 3, the energy is,

${\mathrm{E}}_2=\frac{4{\mathrm{h}}^2}{8{\mathrm{mL}}^2}$and${\mathrm{E}}_3=\frac{4{\mathrm{h}}^2}{9{\mathrm{mL}}^2}$

So,

$$\begin{gathered} ∆\mathrm{E}={\mathrm{E}}_3-{\mathrm{E}}_2 \\ =\frac{5{\mathrm{h}}^2}{8{\mathrm{mL}}^2} \end{gathered}$$

Substitute the value of $∆\mathrm{E}$in equation (ii) and solve for the wavelength,

$$\begin{gathered} \frac{5{\mathrm{h}}^2}{8{\mathrm{mL}}^2}=\frac{\mathrm{hc}}{\mathrm{\lambda }} \\ \mathrm{\lambda }=\frac{8{\mathrm{mcL}}^2}{5\mathrm{h}} \\ =\frac{3}{5}\left(19.2\mathrm{\mu m}\right) \\ =11.5\mathrm{\mu m} \end{gathered}$$

Thus, the wavelength from this transition is 11.5 µm.