# 96: The de Broglie-Bohr Model for the Hydrogen Atom - Version 3

$$\lambda = \frac{h}{mv}$$ de Broglie's hypothesis that matter has wave-like properties.

$$n \lambda = 2 \pi r$$ The consequence of de Broglie's hypothesis; an integral number of wavelengths must fit within the circumference of the orbit. This introduces the quantum number, n, which can have values 1, 2, 3...

$$mv = \frac{nh}{2 \pi r}$$ Substitution of the first equation into the second equation reveals that linear momentum is quantized.

$$T = \frac{1}{2} mv^2 = \frac{n^2h^2}{8 \pi^2 m_e r}$$ If momentum is quantized, so is kinetic energy.

$$E = T + V = \frac{n^2h^2}{8 \pi^2 m_e r^2} - \frac{q^2}{4 \pi \varepsilon o_0 r}$$ Which means that total energy is quantized.

Below the ground state energy and orbit radius of the electron in the hydrogen atom is found by plotting the energy as a function of the orbital radius. The ground state is the minimum in the curve.

Fundamental constants: electron charge, electron mass, Planck's constant, vacuum permitivity.

$\begin{matrix} q = 1.6021777 (10)^{-19} \text{coul} & m_e = 9.10939 (10)^{-31} \text{kg} \\ h = 6.62608 (10)^{-34} \text{joule sec} & \varepsilon_0 = 8.85419 (10)^{-12} \frac{ \text{coul}^2}{ \text{joule m}} \end{matrix}$

Conversion factors between meters and picometers and joules and atto joules.

$\begin{matrix} pm = 10^{-12} m & \text{ajoule} = 10^{-18} \text{joule} & eV = 1.602177 (10)^{-19} \text{joule} \end{matrix}$

Setting the first derivative of the energy with respect to r equal to zero, yields the optimum value of r.

$\begin{matrix} \frac{d}{dr} \left( \frac{n^2h^2}{8 \pi^2 m_e r^2} - \frac{q^2}{4 \pi \varepsilon_0 r} \right) = 0 & \text{has solution(s)} & n^2 h^2 \frac{ \varepsilon_0}{q^2 \pi m_e} \end{matrix}$

Substitution of this value of r back into the energy expression yields the energy gives the energy of the hydrogen atom in terms of the quantum number, n, and the fundamental constants.

$\begin{matrix} E = \frac{n^2h^2}{8 \pi^2 m_e r^2} - \frac{q^2}{4 \pi \varepsilon_0 r} & \text{by substitution, yields} & E = \frac{-1}{8 n^2h^2} \frac{m_e}{ \varepsilon_0^2} q^4 \end{matrix}$

Calculate the allowed energy levels for the hydrogen atom: n = 1 .. 5

$\begin{matrix} E_n = \frac{-1}{8 n^2 h^2} \frac{m_e}{ \varepsilon_0^2} q^4 & \frac{E_n}{ \text{ajoule}} = \begin{pmatrix} -2.18 \\ -0.545 \\ -0.242 \\ -0.136 \\ -0.087 \end{pmatrix} & \frac{E_n}{eV} = \begin{pmatrix} -13.606 \\ -3.401 \\ -1.512 \\ -0.85 \\ -0.544 \end{pmatrix} \end{matrix}$

Prepared by Frank Rioux.