$$\begin{equation} \frac{1}{\lambda} = R_H \left( \frac{1}{m^2} - \frac{1}{n^2} \right) \end{equation}$$
If we let $m = 2$ in the above equation, then setting $n = 3,4,\ldots,6$ yields the following wavelengths for the expected emissions.
n | $\lambda$ [nm] |
---|---|
3 | 656.3 |
4 | 486.1 |
5 | 434.0 |
6 | 410.2 |
7 | 397.0 |
Starting with Coulomb's Law and $F = ma$: $$\begin{equation} -\frac{(Ze)e}{4 \pi \epsilon_0 r^2} = - \frac{m_e v^2}{r} \end{equation}$$ Then taking the kinetic energy of the orbiting particle: $$\begin{equation} K = \frac{1}{2}m_e v^2 = \frac{Ze^2}{8 \pi \epsilon_0 r} \label{eq:kineticenergyelectron} \end{equation}$$ The potential energy is $$\begin{equation} U = - \frac{Ze^2}{4 \pi \epsilon_0 r} \end{equation}$$ Summing the kinetic and potential to get the total energy: $$\begin{equation} E = K + U = \frac{Ze^2}{8 \pi \epsilon_0 r}-\frac{Ze^2}{4 \pi \epsilon_0 r} = -\frac{Ze^2}{8 \pi \epsilon_0 r} \end{equation}$$
The Bohr model suggests that instead of the total energy being capable of taking on any value, instead it can only take on quantized integer multiples of a constant value. Or, in terms of the angular momentum: $$\begin{equation} L = m_e v r = \frac{n h}{2 \pi} = n \hbar \end{equation}$$ Solving for $v$ and combining this with the kinetic energy from equation \eqref{eq:kineticenergyelectron} we'll obtain for the radius, $r$: $$\begin{equation} r_n = \frac{4 \pi \epsilon_0 \hbar^2}{Z e^2 m_e} n^2 = \frac{5.29 \times 10^{-11} \; \textrm{m}}{Z}n^2 \end{equation}$$ Here, $n$ is called a quantum number and can take integer values starting with 1. $Z$ is the number of protons in the nucleus. If $n = 1$ and $Z = 1$, which corresponds to the lowest energy level of the hydrogen atom, $$\begin{equation} r_\textrm{Bohr} =0.529 \times 10^{-11} \; \textrm{m} = 0.529 \unicode{x212B} \end{equation}$$ ($1 \unicode{x212B} = $ one tenth of a nanometer or $1 \times 10^{-10} \; \textrm{m}$)
$E_1$ is the lowest energy of the hydrogen atom: -13.6 eV.
Now, if we are interested in the difference between two energies levels of an electron: $$\begin{equation} \Delta E = E_\rm{high} - E_\rm{low} = E_\rm{photon} \end{equation}$$
Find the wavelength of a photon emitted during a transition from $n = 3$ to $n=2$ for an electron in a hydrogen atom.
The energy of a single, reddish, photon: $$\begin{equation} E_\textrm{photon} = h\nu = \frac{hc}{\lambda} \simeq \frac{1240 \; \textrm{eV nm}}{700 \; \textrm{nm}} = 1.77 \; \textrm{eV} \end{equation}$$ is about half that of a blue photon: $$\begin{equation} E_\textrm{photon} = h\nu = \frac{hc}{\lambda} \simeq \frac{1240 \; \textrm{eV nm}}{400 \; \textrm{nm}} = 3.10 \; \textrm{eV} \end{equation}$$