Question

A hydrogen atom initially in its ground level absorbs a photon, which excites the atom to the n=3 level. Determine the wavelength and frequency of the photon.

Short answer

The wavelength and frequency of the photon are$1.02\times {10}^{-7}\mathrm{m}$and $2.92\times {10}^{15}\mathrm{Hz}$respectively.

Step-by-step solution

Define the energy.

The energy $\mathit{E}$of a photon is inversely proportional to the wavelength $\mathit{\lambda }$of photon and directly proportional to the frequency $\mathit{f}$of wave

$$\begin{gathered} \mathit{E}\mathbf{=}\mathit{h}\mathit{f} \\ \mathit{E}\mathbf{=}\frac{\mathbf{h}\mathbf{c}}{\mathbf{\lambda }} \end{gathered}$$

Where, the value of Planck’s constant $\mathit{h}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{26}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{34}\mathbf{}}\mathbf{J}\mathbf{.}\mathbf{s}$.

In the Bohr model for the hydrogen atom the equations energy${\mathit{E}}_{\mathbf{n}}$ of the $\mathit{n}\mathit{t}\mathit{h}$orbit is:

${\mathit{E}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{-}\mathbf{13}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{e}\mathbf{V}}{{\mathbf{n}}^{\mathbf{2}}}$

Determine the wavelength and frequency of the photon.

The energy of the ground level is:

$$\begin{gathered} E=\frac{-13.6}{1.0} \\ =13.60\mathrm{eV} \end{gathered}$$

The energy of the $n=3$level is:

$$\begin{gathered} {\mathrm{E}}_3=\frac{-13.60}{{\left(3.0\right)}^2} \\ =-1.51\mathrm{ev} \end{gathered}$$

The frequency of the photon is:

$$\begin{gathered} \mathrm{f}=\frac{{\mathrm{E}}_2-{\mathrm{E}}_1}{\mathrm{h}} \\ =\frac{-1.51-\left(-13.60\right)}{4.136\mathrm{x}{10}^{-15}} \\ =2.92\mathrm{x}{10}^{15}\mathrm{Hz} \end{gathered}$$

The wavelength of the photon is:

$$\begin{gathered} \mathrm{\lambda }=\frac{\mathrm{c}}{\mathrm{f}} \\ =\frac{3\mathrm{x}{10}^6}{2.92\mathrm{x}{10}^{15}} \\ =1.02\mathrm{x}{10}^7\mathrm{m} \end{gathered}$$

Hence, the wavelength and frequency of the photon are $1.02\mathrm{x}{10}^7\mathrm{and}2.92\mathrm{x}{10}^{15}\mathrm{Hz}$ respectively.