1924, de Broglie: Electrons act like standing waves in the orbit, and its oscillation wavelength is:
\[ \lambda = \frac{h}{mv} \]
\(h\): Planck constant (\(\approx 6,626.10^{-34}\))
\(m, v\): mass and velocity of the electron

But, if the wave is standing, then a whole number of wavelengths must fit within the orbit's circumference. The first harmonic (\(n=1\)) has 1 oscillation, the second (\(n=2\)) has 2 and so on.
\[\therefore\] \[number\;of\;wavelengths = circumference\] \[1\lambda_1=2 \pi r_1\] \[2\lambda_2=2 \pi r_2\] \[3\lambda_3=2 \pi r_3\] \[\vdots\] \[n\lambda_n=2 \pi r_n\]
Applying the de Broglie wavelength:
\[n\left( \frac{h}{mv_n} \right)=2 \pi r_n\] \[{\color{gray} \frac{nh}{mv_n}=2 \pi r_n}\] \[{\color{gray} \frac{nh}{\color{red}mv_n r_n}=2 \pi}\]
Since angular momentum \(L=mvr\),
\[{\color{gray} \frac{nh}{\color{red}L}=2 \pi}\]
Where \(L\) is now the electron's angular momentum.
Solving for \(L\):
\[{\color{gray}nh=2 \pi L}\] \[{L=\frac{nh}{2 \pi}}\]
Which is Bohr's postulate!
Another thing we can find from this is the orbit's radius (\(r_n\))!
\[{\color{gray} \frac{nh}{mv_n}=2 \pi r_n}\] \[r_n=\frac{nh}{2 \pi mv_n}\]
But we don't know the electron's velocity (\(v_n\))...
An useful analogy is with planetary orbits, or with artificial satellites:
For it to stay in a circular orbit, the centripetal force must equal the gravitational force between the Earth and the satellite:
\[F_c=F_g\]
Applying the same logic to the eletron and the proton of the hydrogen atom, the centripetal force must equal instead the electrostatic force between them, as given by Coloumb's law:
\[|F_e|=k_e \frac{|Qq|}{d²}\]
\(Q, q\): charge magnitudes
\(d\): distance between the charges (here it's the orbit radius)
\(k_e\): Coloumb constant (\(\approx 8,988.10^9\))

We can further define \(k_e\) as:
\[|F_e|=\frac{1}{4 \pi \varepsilon_0} \frac{|Qq|}{d²}\]
\(\varepsilon_0\): vacuum electric permittivity

Applying the centripetal force equality:
\[F_c=F_e\] \[\frac{mv²_n}{r_n}=\frac{1}{4 \pi \varepsilon_0}\frac{|e\cdot -e|}{r²_n}\] \[{\color{gray} mv²_n=\frac{e²}{4 \pi \varepsilon_0 r_n}}\] \[{\color{gray} v²_n=\frac{e²}{4 \pi \varepsilon_0 r_n m}}\] \[v_n=\frac{e}{\sqrt{4 \pi \varepsilon_0 r_n m}}\]
Hm, now there's an annoying square root...
Well, we'll apply this velocity equation on the previous radius equation (inverted because it's dividing)...
\[{\color{gray}r_n=\frac{nh}{2 \pi m{\color{red}v_n}}}\] \[{\color{gray}r_n=\frac{nh}{2 \pi m}} {\color{red}\frac{\sqrt{4 \pi \varepsilon_0 r_n m}}{e}} \]
And square both sides so the square root goes away!
\[{\color{gray}r_n{\color{red}²}=\frac{n²h²}{{\color{red}4} \pi{\color{red}²} m{\color{red}²}}\frac{{\color{red}4 \pi} \varepsilon_0 {\color{red}r_n m}}{e²}}\] \[r_n=\frac{n² h² \varepsilon_0}{\pi me²}\]
Now we have a way to feasibly calculate the radius!
\[{\color{gray} \frac{h² \varepsilon_0}{\pi me²}=0.529 \mathring{A}\;(10^{-10}m)}\]
so for any value of \(n\):
\[r_n=n² \cdot 0.529\mathring{A}\] \[r_1=0.529\mathring{A}\] \[r_2=4r_1=2.116\mathring{A}\] \[r_3=9r_1=4.761\mathring{A}\] \[\vdots\]
The radius for \(n=1\) is called the Bohr radius, the smallest radius of a hydrogen atom!
To figure out the energy of the electron at different levels, we use:
\[E_n=K_n+U_n\]
\(K_n, U_n\): kinetic and potential energies

Where:
\[U_n=-k_e\frac{Qq}{r}\] \[=\frac{1}{4 \pi \varepsilon_0}\frac{-e²}{r}\] \[K_n=\frac{m{\color{red}v²}}{2}\]
Replacing the velocity with the equation we found:
\[K_n=\frac{1}{2}{\color{green}m}{\color{red}\frac{e²}{4 \pi \varepsilon_0 {\color{green}m}r}}\]
Rearranging some terms, you might tell some similarities between \(U_n\) and \(K_n\):
\[K_n=\frac{1}{2}\left[ \frac{1}{4 \varepsilon_0}\frac{e²}{r_n} \right]\] \[U_n=-\left[ \frac{1}{4 \varepsilon_0}\frac{e²}{r_n} \right]\]
So we can rearrange \(E_n=K_n+U_n\) as:
\[E_n=\left[ \frac{1}{4 \varepsilon_0}\frac{e²}{r_n} \right] \left( \frac{1}{2}-1\right)\] \[E_n=-\frac{1}{2}\left[ \frac{1}{4 \varepsilon_0}\frac{e²}{\color{red}r_n} \right]\]
And replacing \(r_n\) with our equation (inverted again):
\[E_n = \frac{-e²}{8 \pi \varepsilon_0} \cdot {\color{red}\frac{\pi me²}{n²h²\varepsilon_0}}\] \[E_n = \frac{-\pi me⁴}{8\pi \varepsilon²_0 h² n²}\]
Once again, we can feasibly calculate the total energy for any n!
\[\frac{-\pi me⁴}{8\pi \varepsilon²_0 h²}\approx -13,6eV\] \[E_n=\frac{-13,6eV}{n²}\] \[E_1=13,6eV\] \[E_2=\frac{-13,6eV}{4}=-3.4eV\] \[E_3=\frac{-13,6eV}{9}=-1.51eV\] \[\vdots\]
And thanks to the similarities between the equations for \(E_n\), \(K_n\) and \(U_n\):
\[U_n=2E_n=\frac{-27,2eV}{n²}\] \[K_n=-E_n=\frac{+13,6eV}{n²}\]