blackbody radiation & planck’s hypothesis a blackbody is any object that absorbs all light...
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Blackbody Radiation & Planck’s HypothesisA 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
Consider heating blackbodies to various temperatures and recording intensity of radiation at differing frequenciesAt both low and high freq.
there is very little radiationThe rad. Peaks at an
intermediate freq.This distribution holds true
regardless of the materialNote: As temp. increases –
area under curve increasesThis represents total energy
As temp. increases – peak moves to higher frequency
The temperature
therefore indicates its emitted color and vice versaWe 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
Planck’s Quantum HypothesisAttempts to explain
blackbody radiation using classical physics failed miserablyAt 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
German physicists Max
Planck diligently tried to solve this issueHe “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
E α fInserting a constant, h
E = n h fWhere n = number of
packets and h = planck’s constant
h = 6.63 x 10-34 J • sOne of our fundamental
constants of natureThis tells us that energy
can only change in quantum jumps, a very tiny amount not experienced everyday
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 neededAs 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
Photons & the Photoelectric EffectPlanck 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
Each photon has energy based on its frequency E = n h f
A beam of light can be thought of as a beam of particlesMore intense =
more particlesSince each photon
have small amounts of energy, there must be tremendous numbers of them
Einstein applied this model to the photoelectric effect issueLight hitting the surface of metals can cause
electrons to be ejectedThe effect could not be explained using the wave
theory of lightWe can determine the number of e ejected by
connecting the apparatus to a simple circuit
The minimum amount
of energy needed to eject e = work function, W0
Metal dependentUsually a few eVIf an e is given
energy by light that exceeds W0, the additional amount goes into kinetic energy of e
Kmax = E – W0
Classical physics predicts 1. light of any
frequency should eject e as long as intensity is high enough
2. The K of e should increase with intensity
These do not agree
with experiment:1. There is a
minimum frequency required – the cutoff frequency, f0
If f < f0 no e regardless of the intensity
2. The Kmax of e depends only on the frequency
Increasing intensity about f0 only increases the number of e
Both of these are explained using the photon model of light1. Changing intensity
only changes the number of photons
2. E is ejected only if the photon has sufficient energy (at least equal to the work function)
The is the cutoff frequency, f0
If f > f0, the e leaves
metal with some KIf f < f0, no e are ejected
regardless of intensitySince 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
Photons & the Photoelectric EffectQuantization of light – Albert
Einstein (1905)Based on properties of EM
wavesEmitted radiation should be
quantizedQuantum (packet of light) –
photonEach photon has energy E = h fLittle bundles of light energy
Connection between wave & particle nature of light
Einstein used this to explain the photoelectric effect
Certain metallic materials are photosensitiveLight striking material emits electrons (e)
The radiant energy supplies the work necessary to free the e – photoelectrons
When photocell is illuminated with monochromatic light, characteristic curves are obtained
Photocurrent until a saturation current is reachedAll emitted e reach anode voltage has no effect on current
Classically: > the intensity, the > energy of eK of e can be tested by reversing voltageOnly e with enough K (eV) make it to the
negative plate & contribute to the currentAs voltage , then is made negative, current At some voltage V0, the stopping potential, no
current will flow
The max K (Kmax) is
related to stopping potentialeV = the work
needed to stop e Kmax = eV0
When f of light is varied, the Kmax is found to depend linearly on fNo photoemission is
observed below cutoff frequency, f0
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 = hfThe e absorb whole
photon or nothing
Since e are bound by attractive forces, work must be done
Conservation of energyhf = 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
Least tightly bound
will have maximum KEnergy 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
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
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