Materials in Electronics/The Hydrogen Atom

The hydrogen atom is the simplest possible atom, comprising a single proton and a single electron.

To determine the wavefunction of the atom's electron, we must solve the 3D Schrödinger wave equation with the Coulomb potential of the single proton:


 * $$- \frac{\hbar}{2m} \left( \frac{d^2}{dx^2} + \frac{d^2}{dy^2} + \frac{d^2}{dz^2} \right) \psi \left( x,y,z \right) + V \left( x,y,z \right)\psi \left( x,y,z \right) = E \psi \left( x,y,z \right) $$
 * $$ V \left( x,y,z \right) = V \left( r \right) = - \frac{e^2}{4 \pi \epsilon_0 r }$$

We re-express the wavefunction in spherical co-ordinates:
 * $$\psi(x,y,z) = \psi(r, \theta, \phi )\,$$

If there is no angular dependence, we can simple solve for the radial component, r.


 * $$\frac{d^2}{dx^2} + \frac{d^2}{dy^2} + \frac{d^2}{dz^2} = \frac{d^2}{dr^2} +\frac{2}{r} \frac{d}{dr}$$

The Schrödinger equation for the radial solution of the hydrogen atom is then:
 * $$- \frac{\hbar}{2m} \left( \frac{d^2}{dr^2} + \frac{2}{r}\frac{d}{dr} \right) \psi \left( r\right) - \frac{e^2}{4 \pi \epsilon_0 r }\psi \left( r \right) = E \psi \left( r \right) $$

Ground State
For the lowest energy state, this differential equation can be solved by the following trial solution:
 * $$\psi(r) = \psi_0 e^{-ar}\,$$

Substituting in, we get:


 * $$- \frac{\hbar}{2m} \left( a^2 e^{-ar} - \frac{2 a e^{-ar} }{r} \right) - \frac{e^2}{4 \pi \epsilon_0 r } e^{-ar} = E e^{-ar}  $$
 * $$- \frac{\hbar}{2m} \left( a^2 - \frac{2 a }{r} \right)  - \frac{e^2}{4 \pi \epsilon_0 r }  = E   $$

The coefficents of r0 and r−1 must be separately satisfied:


 * $$- \frac{ \hbar^2 a^2 }{2m} = E$$
 * $$- \frac{ \hbar^2 a }{m} - \frac{e^2}{4 \pi \epsilon_0} = 0$$

We therefore have values for a and E:


 * $$a = \frac{e^2 m}{4 \pi \hbar^2 \epsilon_0} \approx \frac{1}{0.05 \mbox{ nm} } $$
 * $$E = - \frac{m e^4 } {32 \pi^2 \epsilon_0^2 \hbar^2} \approx 13.8 \mbox{ eV}$$

Higher states
Higher energy levels have a more complex wavefunction:


 * $$\psi_n(r) = \psi_0 e^{-a_nr} L_n(r)\,$$

where Ln(r) is the Laguerre polynomial of order n. It is no important here how this result was reached - for a more thorough explanation, see the Quantum Mechanics Wikibook. The associated energies are:


 * $$E_n = - \frac{m e^2} {32 n^2 \pi^2 \epsilon_0^2 \hbar^2} \approx - \frac{13.6}{n^2} \mbox{ eV}, \quad \quad n=1,2,3,\ldots$$

The energy levels become closer and closer as n increases and the energy approaches zero. n = ∞ corresponds to the vacuum level at which the electron is free of the proton. Electrons can move between the states by absorbing (to move up) or emitting (to move down) the correct amount of energy.

Spherical orbitals
For every n, a spherical solution exists, in which the wavefunction is invariant with angle. These spherical orbitals are known as the s-orbitals.

Non-spherical orbitals
The complete solution for the wavefunction has four quantum numbers:
 * $$\psi_{n,l,m,s}(r,\theta, \phi), \quad l = 0, 1, \ldots, n-1, \quad m = -l, \ldots, +l, \quad s = -\frac{1}{2}, +\frac{1}{2}$$

For example, when n = 2 and l = 1 there are three solutions corresponding to m = −1,0,1. These are known as the p-orbitals. Up to six electrons can be held in the p-orbitals - two for each solution, one spin-up and one spin-down.