Question

a. Derive an expression for ${\mathrm{\lambda }}_{2\to 1}$, the wavelength of light emitted by a particle in a rigid box during a quantum jump from $\mathrm{n}=2\mathrm{to}\mathrm{n}=1.$

b. In what length rigid box will an electron undergoing a $2\to 1$ transition emit light with a wavelength of $694\mathrm{nm}$? This is the wavelength of a ruby laser

Short answer

a. $\mathrm{\lambda }=\frac{8\mathrm{mc}{\mathrm{L}}^2}{3\mathrm{h}}$

b.$\mathrm{E}=1.7867\mathrm{eV}$

Step-by-step solution

Part (a) Step 1: Given information

We have given,

$\mathrm{n}=2\mathrm{to}1$

we have to find the wavelength.

Simplify

We know that,

$$\begin{gathered} \mathrm{E}=\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}({\mathrm{n}}_2^2-{\mathrm{n}}_1^2) \\ \mathrm{E}=\frac{3{\mathrm{h}}^2}{8{\mathrm{mL}}^2} \\ \mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }} \\ \mathrm{\lambda }=\frac{8\mathrm{mc}{\mathrm{L}}^2}{3\mathrm{h}} \end{gathered}$$

Part (b) Step 1: Given information

We have given,

wavelength =$694nm$

we have to find the energy.

Simplify

$$\begin{gathered} \mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }} \\ \mathrm{E}=\frac{1240\mathrm{eV}.\mathrm{nm}}{694\mathrm{nm}} \\ \mathrm{E}=1.7867\mathrm{eV} \end{gathered}$$