Bohr atom

Starting from the following postulates, Bohr developed an improved model of the atom which more closely reflected the experimental behavior.

Atoms have certain stable states, and light is emitted or absorbed when the atom moves between higher/lower energy states.
If two stable states have energy $E_1 > E_2$ , then the frequency of the energy emitted in the transition from state $1 \to 2$ is $\nu_{1\to2} = \frac{E_1 - E_2}{h}$ .
For circular orbits, these states are determined by the quantized angular momentum $mvr = \hbar n$

Assuming electrons are in a circular orbit, the centripetal force equals the coulomb attraction $\frac{mv^2}{r} = \frac{Zq^2}{4\pi\epsilon_0r^2}$ , where $Z$ is the number of protons in the nucleus. From this we find $(mvr)^2 = L^2 = \frac{Zq}{4\pi\epsilon_0} mr$ . From the assumption of quantized angular momentum, $n^2 \hbar^2 = \frac{Zq}{4\pi\epsilon_0} mr$ .

This gives us discrete orbits at which the above relation holds, called the Bohr orbits

r_n = \frac{4\pi\epsilon_0}{mZq^2} n^2 \hbar^2.

We can also derive the quantized velocity and energy from the above.

We define the fine structures constant $\alpha = q^2 / 4\pi\epsilon_0 \hbar c$ to simplify these equations, yielding

\begin{align*} r_n &= \frac{\hbar^2 n^2}{mcZ\alpha} \\ v_n &= \frac{Z\alpha c}{n} \\ E_n &= -\frac12 mc^2 \left(\frac{Z\alpha}{n}\right)^2. \end{align*}

Spectral lines

The energy required for an electron to move from energy level $m$ to $n$ is given by $E_n - E_m$ . When an electron moves from a higher to a lower energy level, the energy lost is emitted in the form of a single photon, of frequency $\nu_{m\to n}=\frac{E_m -E_n}{h} = R_\infty \left( \frac{1}{m^2} - \frac{1}{n^2} \right)$ where $R_\infty \approx 3.25 \cdot 10^{15} \,\mathrm{Hz}$ is the Rydberg constant.

The plot below shows the energy level transitions in hydrogen. Free electrons have $E=0$ , while electrons at the lowest energy level $n=1$ have $E=-13.6 \eV$ .

The arrows correspond to energy level transitions, each of which has a corresponding photon emission frequency.

Angular momentum quantization

Consider the de Broglie wave of the electron. In order for the electron to actually “exist” at a certain radius, the wave function needs to constructively interfere—otherwise, the probability of finding the electron there would go to zero. Thus the circumference of the orbit needs to be a multiple of the wavelength.

Mathematically this gives us the condition

\begin{align*} 2\pi r &= n \lambda_\dB = n \frac hp = n \frac{h}{mv} \\ mvr &= n \frac{h}{2\pi} = \boxed{ n \hbar = L }. \end{align*}

From which we get the quantization of angular momentum.