Absorption and emission

Bohr could now precisely describe the processes of absorption and emission in terms of electronic structure. According to Bohr's model, an electron would absorb energy in the form of photons to get excited to a higher energy level as long as the photon's energy was equal to the energy difference between the initial and final energy levels. After jumping to the higher energy level—also called the excited state—the excited electron would be in a less stable position, so it would quickly emit a photon to relax back to a lower, more stable energy level. 

 

The energy levels and transitions between them can be illustrated using an energy level diagram, such as the example above showing electrons relaxing back to the $n=2$n=2n, equals, 2 level of hydrogen. The energy of the emitted photon is equal to the difference in energy between the two energy levels for a particular transition. The energy difference between energy levels $n_{high}$n​high​​n, start subscript, h, i, g, h, end subscript and $n_{low}$n​low​​n, start subscript, l, o, w, end subscript can be calculated using the equation for $E(n)$E(n)E, left parenthesis, n, right parenthesis from the previous section:

$\begin{aligned} \Delta E &= E(n_{high})-E(n_{low}) \\ \\ &=\left( -\dfrac{1}{{n_{high}}^2} \cdot 13.6\,\text{eV} \right)-\left(-\dfrac{1}{{n_{low}}^2} \cdot 13.6\,\text{eV}\right) \\ \\ &= \left(\dfrac{1}{{n_{low}}^2}-\dfrac{1}{{n_{high}}^2}\right) \cdot 13.6\,\text{eV} \end{aligned}$​ΔE​​​​​​​=E(n​high​​)−E(n​low​​)​=(−​n​high​​​2​​​​1​​⋅13.6eV)−(−​n​low​​​2​​​​1​​⋅13.6eV)​=(​n​low​​​2​​​​1​​−​n​high​​​2​​​​1​​)⋅13.6eV​​

Since we also know the relationship between the energy of a photon and its frequency from Planck's equation, we can solve for the frequency of the emitted photon:

$\begin{aligned} h\nu &=\Delta E = \left(\dfrac{1}{{n_{low}}^2}-\dfrac{1}{{n_{high}}^2}\right) \cdot 13.6\,\text{eV} ~~~~~~~~~~~~\text{Set photon energy equal to energy difference}\\ \\ \nu &= \left(\dfrac{1}{{n_{low}}^2}-\dfrac{1}{{n_{high}}^2}\right) \cdot \dfrac{13.6\,\text{eV}}{h}~~~~~~~~~~~~~~~~~~~~~~\text{Solve for frequency}\end{aligned}$​hν​​ν​​​=ΔE=(​n​low​​​2​​​​1​​−​n​high​​​2​​​​1​​)⋅13.6eV            Set photon energy equal to energy difference​=(​n​low​​​2​​​​1​​−​n​high​​​2​​​​1​​)⋅​h​​13.6eV​​                      Solve for frequency​​

We can also find the equation for the wavelenth of the emitted electromagnetic radiation using the relationship between the speed of light $\text c$cc, frequency $\nu$ν, and wavelength $\lambda$λlambda:

$\begin{aligned}\text c &=\lambda \nu ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\text{Rearrange to solve for }\nu . \\ \dfrac{\text c}{\lambda}&=\nu=\left(\dfrac{1}{{n_{low}}^2}-\dfrac{1}{{n_{high}}^2}\right) \cdot \dfrac{13.6\,\text{eV}}{h}~~~~~~~~~~~~~~\text{Divide both sides by c to solve for }\dfrac{1}{\lambda}.\\ \\ \dfrac{1}{\lambda} &=\left(\dfrac{1}{{n_{low}}^2}-\dfrac{1}{{n_{high}}^2}\right) \cdot \dfrac{13.6\,\text{eV}}{h\text c} \end{aligned}$​c​​λ​​c​​​​​λ​​1​​​​​=λν                                                                  Rearrange to solve for ν.​=ν=(​n​low​​​2​​​​1​​−​n​high​​​2​​​​1​​)⋅​h​​13.6eV​​              Divide both sides by c to solve for ​λ​​1​​.​=(​n​low​​​2​​​​1​​−​n​high​​​2​​​​1​​)⋅​hc​​13.6eV​​​​

 Thus, we can see that the frequency—and wavelength—of the emitted photon depends on the energies of the initial and final shells of an electron in hydrogen.