By the middle of the 19th century it was well known by chemists that excited hydrogen gas emitted a distinct emission spectrum. It was noted that the same lines were always present and that the spacing between these lines became smaller and smaller.
In 1885, the first person to propose a mathematical relationship for these lines was a Swiss high school physics teacher, J.J. Balmer. We now call hydrogen's visible spectrum the Balmer series. Balmer's empirical formula exactly matched the experimentalists' observed wavelengths.
$$\frac{1}{\lambda}=R\biggl(\frac{1}{2^{2}}-\frac{1}{n^{2}}\biggr)~\text{where}\:n\in\bigl(3,4,5,6\bigr)
$$
Where R is called the Rydberg constant and has a well-established value of 1.0974 \times 10$^{7}$ m$^{-1}$ It wasn't until 1913 that Niels Bohr developed a theory of the atom that explained why this formula worked.
In an hydrogen atom, the centripetal force is being supplied by the coulomb force between it and the proton in the hydrogen nucleus.
\begin{aligned}F_{centripetal}&=F_{electrostatic}\\\frac{m\nu_{n}^{2}}{r_{n}}&=\left|k\frac{-e\left(Ze\right)}{r_{n}^{2}}\right|\\m\nu_{n}^{2}&=k\frac{Ze^{2}}{r_{n}}\end{aligned}
Remember that $\mathbf{Z}$ represents the atomic number (the number of protons), that electrons and protons have the same magnitude charge, te, ard that a negative F electrostatic merely means that the electrostatic force is attractive. Also note that the values of $v_n$ of $r_n$ are unknowns in this equation.As a means of evaluating these two unknowns, Bohr first hypothesized that the electron's angular momentum was quantized.
\begin{aligned}
\text{L}&=n\left({\frac{h}{2\pi}}\right) \\
\text{I}\omega&=n\left(\frac{h}{2\pi}\right) \\
mr_{n}^{2}\left(\frac{\nu_{n}}{r_{n}}\right)&=n\biggl(\frac{h}{2\pi}\biggr) \\
mv_{n}r_{n}&=n\left(\frac{h}{2\pi}\right)
\end{aligned}
Upon solving the angular momentum equation for $\mathbf{v}_{n\cdot}$ substituting it into the centripetal force equation yields the following expfession for $r_{n}$\begin{aligned}
m\nu_{\pi}^{2}&=k\frac{ze^{2}}{r_{n}} \\
m\left({\frac{nh}{2\pi mr_{n}}}\right)^{2}&=k\frac{ze^{2}}{r_{n}} \\
m\left(\frac{n^{2}k^{2}}{4\pi^{2}m^{2}r_{n}^{2}}\right)&=k\frac{ze^{2}}{r_{n}} \\
\frac{n^{2}h^{2}}{4\pi^{2}mr_{n}}&=kZe^{2} \\
\text{700}&=\frac{n^{2}h^{2}}{4\pi^{2}kmZe^{2}} \\
r_{R}&=n^{2}\left(\frac{h^{2}}{4\pi^{2}kmZe^{2}}\right)
\end{aligned}For a ground state hydrogen electron,
\begin{aligned}n&=1\:\text{and}\:Z=l\\r_1&=\frac{h^2}{4\pi^2kme^2}\\r_1&=0.53\times10^{-10}\:\text{meters}\end{aligned} or approximately half of an Angstrorn.
Recall that the electric potential energy for an electron would equal
\begin{aligned}
E_{PE} &=qV_{abs} \\
E_{PE}& =-e\left(\frac{k(Ze)}{r_{n}}\right) \\
E_{PE} &=-\frac{k\left(Ze^{2}\right)}{r_{n}}
\end{aligned}
\begin{aligned}
\text{L}&=n\left({\frac{h}{2\pi}}\right) \\
\text{I}\omega&=n\left(\frac{h}{2\pi}\right) \\
mr_{n}^{2}\left(\frac{\nu_{n}}{r_{n}}\right)&=n\biggl(\frac{h}{2\pi}\biggr) \\
mv_{n}r_{n}&=n\left(\frac{h}{2\pi}\right)
\end{aligned}
Upon solving the angular momentum equation for $\mathbf{v}_{n\cdot}$ substituting it into the centripetal force equation yields the following expfession for $r_{n}$\begin{aligned}
m\nu_{\pi}^{2}&=k\frac{ze^{2}}{r_{n}} \\
m\left({\frac{nh}{2\pi mr_{n}}}\right)^{2}&=k\frac{ze^{2}}{r_{n}} \\
m\left(\frac{n^{2}k^{2}}{4\pi^{2}m^{2}r_{n}^{2}}\right)&=k\frac{ze^{2}}{r_{n}} \\
\frac{n^{2}h^{2}}{4\pi^{2}mr_{n}}&=kZe^{2} \\
\text{700}&=\frac{n^{2}h^{2}}{4\pi^{2}kmZe^{2}} \\
r_{R}&=n^{2}\left(\frac{h^{2}}{4\pi^{2}kmZe^{2}}\right)
\end{aligned}For a ground state hydrogen electron,
\begin{aligned}n&=1\:\text{and}\:Z=l\\r_1&=\frac{h^2}{4\pi^2kme^2}\\r_1&=0.53\times10^{-10}\:\text{meters}\end{aligned} or approximately half of an Angstrorn.
Bohr's second hypothesis in his model was that an eiectron only loses or reieases energy (and therefore a photon) when it goes trough de-excitation or drops from a higher energy state to a lower energy state. In order to determine the energy lost by the electron, an expression for an electron's total energy has to be developed.
Recall that the electric potential energy for an electron would equal
\begin{aligned}
E_{PE} &=qV_{abs} \\
E_{PE}& =-e\left(\frac{k(Ze)}{r_{n}}\right) \\
E_{PE} &=-\frac{k\left(Ze^{2}\right)}{r_{n}}
\end{aligned}
By extending the centripetal force relationship, an expression can also be derived for the electron's kinetic energy
\begin{aligned}
F_{centripetal}& =F_{electrostatic} \\
\frac{mv_{n}^{2}}{r_{n}}& =\left|k\frac{-e(Ze)}{r_{n}^{2}}\right| \\
m\nu_{n}^{2}& =k\frac{Ze^{2}}{r_{n}} \\
\frac{1}{2}m\nu_{n}^{2}& =k\frac{Ze^{2}}{2r_{n}} \\
KE& =k\frac{Ze^{2}}{2r_{n}}
\end{aligned}
F_{centripetal}& =F_{electrostatic} \\
\frac{mv_{n}^{2}}{r_{n}}& =\left|k\frac{-e(Ze)}{r_{n}^{2}}\right| \\
m\nu_{n}^{2}& =k\frac{Ze^{2}}{r_{n}} \\
\frac{1}{2}m\nu_{n}^{2}& =k\frac{Ze^{2}}{2r_{n}} \\
KE& =k\frac{Ze^{2}}{2r_{n}}
\end{aligned}
Thus, the total energy, \(E_n\) , of an electron equals
\begin{aligned}E_{n}& =E_{PE}+KE \\
&=-\frac{k\left(Ze^{2}\right)}{r_{n}}+k\frac{Ze^{2}}{2r_{n}} \\
&=-k\frac{Ze^{2}}{2r_{n}}
\end{aligned}
In this equation, notice that the total energy is negative. This is interpreting as meaning that the electron is trapped in
an energy well about the nucleus; that is, it would take the addition of energy to ionize or free the electron.
Substituting in the value for r$_{1}$ into this total energy expression yields a ground state energy of 2.18 x 10$^{-18}$ Joules or \(-13.6 \text{eV}\) for a hydrogen atom. Using the fact that $r_n=n^2r_1$
\begin{aligned}r_n&=n^2\bigg(\frac{h^2}{4\pi^2kmZe^2}\bigg)\\r_n&=n^2r_1\end{aligned}
we can now generated the first four energy levels for hydrogen.
\begin{aligned}&E_1 \text{= -13.6 eV} \\
&E_2 =E_1/2^2=-3.4\mathrm{~eV} \\
&E_3 =E_1/3^2=-1.51\mathrm{~eV} \\
&E_4 =E_1/4^{2}=-0.85\mathrm{eV}
\end{aligned}
Bohr's second hypothesis combined with Planck's formula for quantized energy \(E = hf\) will now allow us to derive Balmer's equation. Remember that the energy released by the electron during de-excitation equals the energy of the emitted photon.
Let's begin by assuming that an electron is falling from \(E_j\) , a
high energy state, to \(E_i\) , a lower energy state.
\begin{aligned}\Delta E& =E_{j}-E_{i} \\
\Delta E& =-k\frac{Ze^{2}}{2r_{j}}-\left(-k\frac{Ze^{2}}{2r_{i}}\right) \\
\Delta E& =-k\frac{Ze^{2}}{2}\left(\frac{1}{r_{j}}-\frac{1}{r_{i}}\right) \\
\text{where}~r_{n}& =n^{2}\left(\frac{h^{2}}{4\pi^{2}kmZe^{2}}\right) \\
\Delta E& =-k\frac{Ze^{2}}{2}\left(\frac{1}{j^{2}\left(\frac{h^{2}}{4\pi^{2}kmZe^{2}}\right)}-\frac{1}{i^{2}\left(\frac{h^{2}}{4\pi^{2}kmZe^{2}}\right)}\right) \\
\Delta E& =-k\frac{Ze^{2}}{2}\Bigg(\frac{4\pi^{2}kmZe^{2}}{h^{2}}\Bigg)\Bigg(\frac{1}{j^{2}}-\frac{1}{i^{2}}\Bigg) \\
\Delta E& =-k^{2}\frac{2\pi^{2}mZ^{2}e^{4}}{h^{2}}\Bigg(\frac{1}{j^{2}}-\frac{1}{i^{2}}\Bigg) \\
\Delta E& =k^{2}\frac{2\pi^{2}mZ^{2}e^{4}}{h^{2}}\left(\frac{1}{i^{2}}-\frac{1}{j^{2}}\right) \\
\Delta E_{\text{lost by the electron}}& =E_{\text{photon}}=hf=h\frac{c}{\lambda} \\
h{\frac{c}{\lambda}}& =k^{2}\frac{2\pi^{2}mZ^{2}e^{4}}{h^{2}}\left(\frac{1}{i^{2}}-\frac{1}{j^{2}}\right) \\
\frac{1}{\lambda}& =k^{2}\frac{2\pi^{2}mZ^{2}e^{4}}{ch^{3}}\left(\frac{1}{i^{2}}-\frac{1}{j^{2}}\right) \\
F_{or}Z& =1, \\
\frac{1}{\lambda}& =k^{2}\frac{2\pi^{2}me^{4}}{ch^{3}}\left(\frac{1}{i^{2}}-\frac{1}{j^{2}}\right) \\
\frac{1}{\lambda}& =1.097\times10^{7}\left(\frac{1}{i^{2}}-\frac{1}{j^{2}}\right) \\
\frac{1}{\lambda}& =R\left(\frac{1}{i^{2}}-\frac{1}{j^{2}}\right)
\end{aligned}
If we let \(i = 2\), and \(j\in\{3,4,5,6\}\) then we have derived Balmer's empirical formula.
