# Bohr Model of the Hydrogen Atom - Energy Levels of the Hydrogen Atom

Why the allowed (stationary) orbits correspond to those for which the orbital angular momentum of the electron is an integer multiple of $\hbar=\frac {h}{2\pi}$? $$L=n\hbar$$ Bohr Quantization rule of the angular momentum that leads to energy levels of the Hydrogen atom: $$E_n=-\frac {Z^2}{1+\frac {m_e}{M_p}} \frac {\Re}{n^2}$$

the question is Why quantum numbers are natural numbers? $$E=nh\nu$$ $$n=0,1,2,3....$$

• Because the constants were picked that way. – Ignacio Vazquez-Abrams May 17 '12 at 19:29
• This is explained in detail in the Wikipedia page on Bohr model and "correspondence principle", what is unclear? – Ron Maimon May 17 '12 at 22:17
• You may want to also see en.wikipedia.org/wiki/De_Broglie_hypothesis. – tmac May 17 '12 at 22:51
• Photons can't be in fragmented parts, so n can't be decimals. Thus, n is always whole number. – Immortal Player Dec 10 '13 at 9:28

The Bohr model wasn't the right theory of all of atomic physics but it described the levels of the Hydrogen atom correctly, due to a mathematical coincidence related to this solvable mathematical problem in the full quantum mechanics.

The integrality conditions of the Bohr model were ad hoc – chosen so that the energy levels as seen in the Hydrogen absorption/emission spectra could be matched – but the most justified starting point to derive them was the Sommerfeld-Wilson quantization condition $$\int_0^T p_r\cdot dq_r = nh$$ where the integral of $p\,dq$ goes over one orbital period. In this form, it is analogous to the statement in the full quantum mechanics – that replaced the Bohr model in the mid 1920 – that the phase space (space parameterized by the positions $q$ and momenta $p$) is composed of cells whose area (or volume) is equal to multiples of $h=2\pi\hbar$ (or powers of $h$, if there are many coordinates). The orbit encircles an area in the phase space and the area should be quantized.

By a coincidence, this is also pretty much equivalent to the quantization of the angular momentum, $L=n\hbar$. In quantum mechanics, similar conditions hold but for slightly different reasons and the quantization of the angular momentum allows half-integral values, too: $J=n\hbar/2$ where $n$ is integer according to quantum mechanics.

One must separate the explanations in the Bohr model from those in the proper quantum mechanics; they're inequivalent because the models are inequivalent, too. And it doesn't make too much sense to think about the origin of the conditions in the Bohr model because the Bohr model is fundamentally not the right theory as we know today.

In the full quantum mechanics, one may encounter several "quantization" facts with the quantum proportional to $h$ or $\hbar$ or $\hbar/2$. All of them have a quantum origin but the detailed explanation is different for each: the quantization of the angular momentum; the elementary cell of the phase space; the unphysical shifts of the action by a multiple of $h$.

• While the Bohr model is not right, the justification of the Bohr model by the correspondence principle is fundamentally correct in the modern understanding too--- the reason for the quantization rule is that the emitted photon should have the classical period at classical motions. – Ron Maimon May 19 '12 at 20:16
• Dear Ron, I am not sure in what sense you may have "fundamentally correct" explanations using theories that are not right. Something about $\int p\,dq$ is quantized in units of $h$ in the right theory but what this quantized quantity is clearly isn't exactly the integral over one period in a wrong Bohr model of the atom so the agreement is vague and a coincidence. – Luboš Motl May 20 '12 at 6:33
• It's not a coincidence. The Bohr model is a wrong picture, but the idea that the level spacing near energy E must be equal to the classical period of the orbit with energy E is correct at large quantum numbers, and must be, because Bohr's correspondence argument is basically sound. If you couple a classical oscillator to EM field weakly, the emitted EM waves classically have frequency equal to the classical orbital frequency $\omega$ (and integer multiples thereof), so if it's photons emitted, the energy levels of the oscillator must be spaced with spacing $h\omega$. – Ron Maimon May 20 '12 at 7:31
• The key point is: energy-level spacing = classical frequency = (2pi/classical-period). This is the identity Bohr uses to find everything else. It must be so for photon coupling to make sense, and it is the reason that the Bohr theory was more than a guess, but a proper seed for Heisenberg to complete to matrix mechanics. – Ron Maimon May 20 '12 at 7:33
• @LubošMotl: Yes, it's true, nobody had a clear idea of the statistical level transitions until Einstein's A,B coefficients, but Bohr's theory reproduces the right semiclassical energy levels (in the large N limit) for any integrable system and this is not a coincidence. The reason is that there is a consistency condition: the level-spacing must be such that photon emission must be possible at the classical frequencies (integer multiples of the orbital frequency). This is Bohr's correspondence principle, and it really is right for the energy levels to leading order in $\hbar$. Read my answer. – Ron Maimon May 26 '12 at 2:47

The basic rule that justified the quantization of angular momentum can be derived as follows, from the correspondence principle. This derivation is heuristic and inexact, and only becomes absolutely correct once you know full quantum mechanics.

Consider an electron orbiting a nucleus of charge 1 (proton, deuteron, or triton) at large radius R with angular momentum L. The classical period is found by equating the centripetal force to the electrostatic attraction of the electron to the nucleus:

$${mv^2\over R} = { e^2\over R^2}$$

Where ${1\over 4\pi\epsilon_0}$ factor is absorbed into the constant e. You find the orbital velocity v from this, so

$$v = \sqrt{ e^2\over m R}$$

Which tells you how long to go around the circle $T={2\pi R\over v}$. So the angular frequency $\omega$ of the orbit

$${2\pi\over T} = \omega = -\sqrt{e^2\over m R^3}$$

The kinetic energy of the orbiting electron is found directly from the centripetal formula:

$${mv^2\over 2} = {e^2\over 2R}$$

The potential energy is:

$$-{e^2\over R}$$

So the total energy is half the negative potential energy

$$- { e^2\over 2R}$$

Classically, this system will radiate electromagnetic waves which are periodic with period T. This means the outgoing radiation has frequency $\omega$. Quantum-mechaically, the orbiting electron can only emit photons with discrete lumps of energy, and this means that the energy can only change in steps of $\hbar\omega$, which is the energy of a photon of frequency $\omega$.

This means that if you have consistent photon emission, the energies must be spaced in discrete energy levels, and the spacing between two adjacent levels at large R is equal to the classical orbital frequency:

$$\Delta E = \hbar \omega = -\hbar\sqrt{e^2\over m R^3}$$

This condition means that if there is an energy level at $E$, there is another energy level at $E-\Delta E$ (where you end up after one photon emission), then another at $E-2\Delta E$ in discrete steps.

This is all semi-classical reasoning, and only really works when the spacing $\Delta E$ is much less than the kinetic energy and the potential energy. The spacing goes to zero as the 3/2 power of the radius, so this approximation is valid for large orbits.

You can figure out the R-spacing between adjacent orbits too

$$E = -{e^2\over 2R}$$ $$dE = {e^2\over 2R^2} dR$$

So the spacing in E translates to a spacing in R (in the approximation that $\Delta E$ and therefore $\Delta R$ are both small so that they approximate the infinitesimal differentials above)

$$\Delta R = {2R^2\over e^2} \Delta E = 2 \hbar \sqrt{R\over m e^2}$$

The change in E and R in each step is complicated, but $\hbar$ has the same units as angular momentum, and you can compute the change in angular momentum when you make a single step:

$$L = mv R = \sqrt{e^2 m R}$$ $$dL = {1\over 2} \sqrt{e^2 m\over R} dR$$

So that

$$\Delta L = {1\over 2} \sqrt{e^2 m \over R} \Delta R = \hbar$$

This is very simple--- the angular momentum is spaced in integer multiples of $\hbar$ at large circular orbits. From this, one can make the plausible guess that this is true for all quantum numbers, large and small, and then the Bohr model follows.

The generalization of this argument to derive the old quantum condition is to consider the period of classical orbits, and to make the energy spacing equal to $\hbar$ times the orbital frequency for a general system. This requirement means that semiclassically:

$$J = \int p dq = 2\pi n \hbar = n h$$

This is shown on the Wikipedia page for correspondence principle. The same quantity J is an adiabatic invariant, it doesn't change under slow deformations of a classical system, and the quantized quantity must have this property, since a slow deformation doesn't have the high frequencies required to make state transitions in quantum mechanics. This argument is summarized on the wikipedia page on the adiabatic invariants.