Blackbody Radiation & Planck’s Hypothesis A blackbody is any object that absorbs all light...

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