1. Blackbody Radiation & Planck’s Hypothesis
A blackbody is any object that absorbs all light incident upon it
Shiny & reflective objects are poor blackbodies
Recall: good absorbers and also good emitters
Ideally we imagine a box with a small hole that very little light (EM radiation) can reflect back out

2. Note: As temp. increases – area under curve increases
Consider heating blackbodies to various temperatures and recording intensity of radiation at differing frequencies
At both low and high freq. there is very little radiation
The rad. Peaks at an intermediate freq.
This distribution holds true regardless of the material
Note: As temp. increases – area under curve increases
This represents total energy
As temp. increases – peak moves to higher frequency

3. The temperature therefore indicates its emitted color and vice versa
We can determine star temperature (surface) by analyzing its color
Red stars are fairly cool, like the bolt shown
But White, or Blue-White stars are very hot
Our sun is intermediate

4. Planck’s Quantum Hypothesis
Attempts to explain blackbody radiation using classical physics failed miserably
At low temps. Prediction & exp match well
At high temps. Classical prediction explodes to infinity
Very different from experimental result
Referred to as the Ultraviolet Catastrophe

5. German physicists Max Planck diligently tried to solve this issue
He “stumbled” upon a mathematical formula that matched the experiment
He then needed to derive the physical formula
The only way was to assume energy (in the form of EM radiation) way quantized
Little “packets” of energy

6.  E α f
Inserting a constant, h
 E = n h f
Where n = number of packets and h = planck’s constant
h = 6.63 x J • s
One of our fundamental constants of nature
This tells us that energy can only change in quantum jumps, a very tiny amount not experienced everyday

7. It does explain the exp. quite well:
Planck was not satisfied and believed (along with other physicists) that it was a purely mathematical solution, not a “real” physical one
It does explain the exp. quite well:
The > f, the > quantum of energy needed
As frequency increased, the amount of energy needed for small jumps increased as well
The object only has a certain amount of energy to supply
Therefore: radiation drops to zero at high frequency

8. Photons & the Photoelectric Effect
Planck believed that the atoms of a blackbody vibrated with discrete frequencies (like standing waves)
But, at the time light was considered a wave therefore no connection
Einstein took the idea of quanta of energy and applied it to light – called photons

9. Each photon has energy based on its frequency  E = n h f
A beam of light can be thought of as a beam of particles
More intense = more particles
Since each photon have small amounts of energy, there must be tremendous numbers of them

10. Einstein applied this model to the photoelectric effect issue
Light hitting the surface of metals can cause electrons to be ejected
The effect could not be explained using the wave theory of light
We can determine the number of e ejected by connecting the apparatus to a simple circuit

11. Classical physics predicts
light of any frequency should eject e as long as intensity is high enough
The K of e should increase with intensity
The minimum amount of energy needed to eject e = work function, W0
Metal dependent
Usually a few eV
If an e is given energy by light that exceeds W0, the additional amount goes into kinetic energy of e
 Kmax = E – W0

12. Both of these are explained using the photon model of light
Changing intensity only changes the number of photons
E is ejected only if the photon has sufficient energy (at least equal to the work function)
The is the cutoff frequency, f0
These do not agree with experiment:
There is a minimum frequency required – the cutoff frequency, f0
If f < f0 no e regardless of the intensity
The Kmax of e depends only on the frequency
Increasing intensity about f0 only increases the number of e

13. If f > f0, the e leaves metal with some K
If f < f0, no e are ejected regardless of intensity
Since energy is that of a photon
 Kmax = hf – W0
Therefore, Kmax depends linearly on frequency
A plot of Kmax for Na & Au shows different cutoff frequencies, but the same slope, h

14. Photons & the Photoelectric Effect
Quantization of light – Albert Einstein (1905)
Based on properties of EM waves
Emitted radiation should be quantized
Quantum (packet of light) – photon
Each photon has energy
 E = h f
Little bundles of light energy
Connection between wave & particle nature of light

15. Einstein used this to explain the photoelectric effect
Certain metallic materials are photosensitive
Light striking material emits electrons (e)
The radiant energy supplies the work necessary to free the e – photoelectrons

16. Photocurrent until a saturation current is reached
When photocell is illuminated with monochromatic light, characteristic curves are obtained
Photocurrent until a saturation current is reached
All emitted e reach anode
 voltage has no effect on current

17. Classically: > the intensity, the > energy of e
K of e can be tested by reversing voltage
Only e with enough K (eV) make it to the negative plate & contribute to the current
As voltage , then is made negative, current 
At some voltage V0, the stopping potential, no current will flow

18. The max K (Kmax) is related to stopping potential
eV = the work needed to stop e
 Kmax = eV0
When f of light is varied, the Kmax is found to depend linearly on f
No photoemission is observed below cutoff frequency, f0

19. Emission begins the instant (~10-9 s) even with low intensity light
Classically, time is required to “build up” energy
Since light can be considered a “bundle of energy”, E = hf
The e absorb whole photon or nothing
Since e are bound by attractive forces, work must be done
Conservation of energy
hf = K + φ
where φ = amount of work (energy) needed to free e
Part of energy of photon “frees” e & the rest is carried away as K

20. Least tightly bound will have maximum K
Does not change energy of individual photons
Photon energy depends on frequency
Below a certain freq. no e are dislodged
When Kmax = 0 the minimum cutoff frequency, f0
Hf0 = Kmax + φ0 = 0 + φ0
 f0 = φ0 / h
Least tightly bound will have maximum K
Energy needed = work function, φ0
hf = Kmax + φ0
Other e require more energy & the K is less
Increasing light intensity, increases # of photons thus increasing # of e

21. Photon has enough energy to free e, but no extra to give it K
Sometimes called threshold frequency
Light below this (no matter how many) will not dislodge e