Tuesday, February 10, 2009


Arthur Compton and Debye both provided in 1922 a very simple mathematical framework for the momentum of these photons with Compton having experimental evidence from firing X-Rays of known frequency into graphite and looking at recoil electrons.

Let E = mc² = hf for a photon, where f is frequency, and "m" is the mass "equivalent" of the photon given they have no "rest mass". (It is important to recognise that stopping a photon to measure its mass eliminates it -so it has no "at rest" mass - crucial in Special Relativity where, to travel at the speed of light, mass would otherwise become infinite.)

Having "rigged" this mass problem,

p = momentum = mc (mass x velocity) = hf/c = E/c = h/l

The experiment shows that X-Rays and electrons behave exactly like ball bearings colliding on a table top using the same 2D vector diagrams. They enter the graphite at one wavelength and leave at a longer wavelength as they have transfered both momentum and kinetic energy to an electron. Momentum and energy are conserved in the collision if we accept the equation above for momentum of light.

When the photon enters at l0 and leaves at l1, its energy has changed from E0 to E1 and momentum from E0/c to E1/c with a change in direction of q. The electron gains Ek = E0 - E1

See for a diagram and more details


The Compton Effect ( a different explanation)

Convincing evidence that light is made up of particles (photons), and that photons have momentum, can be seen when a photon with energy hf collides with a stationary electron. Some of the energy and momentum is transferred to the electron (this is known as the Compton effect), but both energy and momentum are conserved in this elastic collision. After the collision the photon has energy hf' and the electron has acquired a kinetic energy K.

Conservation of energy: hf = hf' + K

Combining this with the momentum conservation equations, it can be shown that the wavelength of the outgoing photon is related to the wavelength of the incident photon by the equation:

Δλ = λ' - λ = (h/mec)(1 - cosq)

The combination of factors h/mec = 2.43 x 10-12 m, where me is the mass of the electron, is known as the Compton wavelength. The collision causes the photon wavelength to increase by somewhere between 0 (for a scattering angle of 0°) and twice the Compton wavelength (for a scattering angle of 180°).


It has problems and examples and demonstration also.

Saturday, February 7, 2009

Schrodinger’s Equation

Quantum mechanics describes the spectra in a much better way than Bohr’s model.

Electron has a wave character as well as a particle character. The wave function of the electron ψ(r,t ) is obtained by solving Schrodinger’s wave equation. The probability of finding an electron is high where | ψ(r,t )|² is greater. Not only the information about the electron’s position but information about all the properties including energy etc. that we calculated using the Bohr’s postulates are contained in the wave function of ψ(r,t).

Quantum Mechanics of the Hydrogen Atom

The wave function of the electron ψ(r,t) is obtained from the Schrodinger’s equation

-(h²/8π²m) [∂²ψ /∂x² + ∂²ψ /∂y² + ∂²ψ/∂z²] - Ze²ψ/4πε0r = E ψ

(x.y,z ) refers to a point with the nucleus as the origin and r is the distance of this point from the nucleus.
E refers to the energy.
Z is the number of protons.

There are infinite number of functions ψ(r,t) which satisfy the equations.

These functions may be characterized by three parameters n,l, and ml.

For each combination of n,l, and ml there is an associated unique value of E of the atom of the ion.

The energy of the wave function of characterized by n,l, and ml depends only on n and may be written as

En = - mZ²e4/8 ε0²h²n²

These energies are identical with Bohr’s model energies.

The paramer n is called the principal quantum number, l the orbital angular momentum quantum number and ml. The magnetic quantum number.

When n = 1, the wave function of the hydrogen atom is

ψ(r) = ψ100 = √(Z³/ π a0²) *(e-r/ a0)

ψ100 denotes that n =1, l = 0 and ml = 0

a0 = Bohr radius

In quantum mechanics, the idea of orbit is invalid. At any instant the wve function is spread over large distances in space, and wherever ψ≠ 0, the presence of electron may be felt.

The probability of finding the electron in a small volume dV is | ψ(r)| ² dV
We can calculate the probability p(r)dr of finding the electron at a distance between r and r+dr from the nucleus.

In the ground state for hydrogen atom it comes out to be

P(r) = (4/ a0)r²e -2r/ a0

The plot of P(r) versus r shows that P(r) is maximum at r = a0 Which the Bohr’s radius.

But when we put n =2, the maximum probability comes at two radii one near r = a0 and the other at r = 5.4 a0. According to Bohr model all electrons should be at r = 4 a0.