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Physics 249 Lecture 3, Sep 10th 2012
Reading: Today Chapter 3, Next time Chapter 4
HW: Due Friday, posted on web site.
1) Discrete nature of radiation
Seeking to understand both the distribution of energy radiated by an ideal black body and
the total energy or power.
∞
Total energy: 𝑈 = ∫0 𝑢(𝜆)𝑑𝜆
The energy density: 𝑢(𝜆) = 𝐸̅ 𝑛(𝜆)
Where n is the number density is as a function of wavelength: 𝑛(𝜆) = 8𝜋𝜆−4
Quantum approach. Planks’s Law:
Plank approached the problem in another way. In his approach he assumed the average
energy at given wavelength was incorrect. Again consider calculating the average energy
starting with the Maxwell-Boltzmann distribution for the energy of the oscillators in a
system with temperature T.
𝑓(𝐸) = 𝐴𝑒 −𝐸/𝑘𝑇
The solution that matches the data is to limit in a discrete way the allowed energies the
oscillators could emit light at and relate the light energy and frequencies or wavelengths.
𝐸𝑛 = 𝑛ℎ𝑓 =
𝑛ℎ𝑐
, 𝑛 = 0,1,2,3 …
𝜆
Relating the light emitted to a fixed frequency made sense since classically light could be
emitted by accelerating (oscillating charges) at the frequency of oscillation. Relating the
frequency (or wavelength) and energy was the key new non-classical idea.
By introducing the wavelength into the Maxwell-Boltzmann distribution by relating
energy and wavelength the behavior of the energy density distribution as a function of
wavelength in the next step can be modified and the integral can be made finite. h is
known as Planck’s constant.
Given this idea you can then calculate information about the distributions to see if it is
consistent with the experimental data.
The average energy:
∞
𝐸̅ = ∫ 𝐸𝑓(𝐸) ⟶ ∑ 𝐸𝑛 𝑓𝑛 =
𝑛=0
∞
𝐸𝑛
∑ 𝐸𝑛 𝐴𝑒 −𝑘𝑇
𝑛=0
∞
𝑐
= ∑ 𝐸𝑛 𝐴𝑒 −𝑛ℎ𝜆𝑘𝑇 =
𝑛=0
ℎ𝑐/𝜆
−1
𝑒 ℎ𝑐/𝜆𝑘𝑇
You now have an average energy for every wavelength. Remember this is the average
energy emitted by an oscillator with a corresponding energy, E.
It is interesting to calculate the behavior of the average energy for each wavelength at
long wavelength.
𝐸̅ =
ℎ𝑐/𝜆
ℎ𝑐/𝜆
1
=
= 𝑘𝑇 ∝
− 1 ℎ𝑐/𝜆𝑘𝑇
𝜆̅
𝑒 ℎ𝑐/𝜆𝑘𝑇
As the temperature goes up the distribution shifts to shorter wavelengths!
The energy density:
8𝜋ℎ𝑐/𝜆5
𝑒 ℎ𝑐/𝜆𝑘𝑇 − 1
𝑢(𝜆) =
which clearly goes down at large wavelengths but also goes to zero as wavelength goes to
zero. This exactly matches the observed distribution after fitting for the unknown h.
The total power
∞
𝑈 = ∫ 𝑢(𝜆)𝑑𝜆 = 𝜎𝑇 4
0
Thus deriving the Stefan Boltzmann constant.
What does the quantized n mean in:
𝐸𝑛 = 𝑛ℎ𝑓 =
𝑛ℎ𝑐
𝜆
, 𝑛 = 0,1,2 …?
It means that the oscillators with energy E can emit light with frequency f, given by E=hf,
n times. In the modern interpretation the oscillating atoms or molecules emit light in
discrete packets known as photons. Quantization of light had been proved, though no one
understood that at the time.
In addition a constant as fundamental as the speed of light had been understood. Planck’s
constant: h = 6.626x10-34 Js.
2) The photoelectric effect:
Depending on the material it takes a certain amount energy to eject an electron from an
atom known as the work function. Classically a wave at any frequency should just build
up energy until it can eject the atom. However, experimentally the energy has to be
larger than the work function to eject an electron. No matter how intense the light beam
if the energy or frequency of the light is too small no electrons will be ejected from the
material.
𝐸 = ℎ𝑓 > 𝜙
To explain this light had to be a discrete particle, a photon, such that only a photon with
sufficient energy can eject an electron.
3) Transition probabilities
Now given the idea that a photon can be absorbed by an electron increasing its energy
lets examine a consequence.
If you have a system with two energy levels E1 and E2 where E2 > E1 then a photon of
energy E = hf = E2 - E1 can cause a transition in the system from energy state 1 to 2.
Similarly the system can decay from energy state 2 to 1 which will happen since it lowers
the systems potential energy. Over a large number of such systems this will happen with
some probability and corresponding lifetime and you can calculate a transition
probability per unit time.
Then you can calculate rates of transition as
Photon emission 2  1: A21*N2
Photon absorption 1  2: B12*n(f)*N1
This starts to set up a simple model that allows you to understand the Quantum
explanation of back body radiation in terms of transitions.
However, it turns out that these two transitions are not enough. You need to add a
process where when the system in energy level 2 is exposed to a photon of the right
frequency it stimulates an emission of a second photon known as stimulated emission.
Stimulated emission 2  1: B21*n(f)*N2
𝑁
Then at equilibrium: 𝑁1 =
2
𝐴21 +𝐵21 𝑛(𝑓)
𝐵12 𝑛(𝑓)
𝑁
Where using the SB distribution 𝑁1 = 𝑒 −𝐸1 /𝑘𝑇 𝑒 𝐸2 /𝑘𝑇 = 𝑒 ℎ𝑓/𝑘𝑇
2
Solving n(f): 𝑛(𝑓) =
𝐴21 /𝐵12
ℎ𝑓
𝑒 𝑘𝑇 −𝐵21 /𝐵12
which for B12 = B21 and the correct relationship between B12 and A21 gives the energy
density relationship derived above.
4) The Compton effect.
The scattering of a photon off an electron transferring some of the photons energy and
momentum to the to the electron.
We will study this in detail when we get to particle physics. Note to solve the problem
you have to apply both quantum mechanics treating the photon as a discrete particle with
energy given by E=hf and relativity in that you have to conserve relativistic energy and
momentum. All of high energy particle scattering interactions have to be treated this way
applying both quantum mechanics and relativity at once.