Problems with Classical Physics. Blackbody Radiation Photoelectric Effect Compton Effect Bohr Model of Atom

The Quantum Gang

Problems with Classical Physics Blackbody Radiation Photoelectric Effect Compton Effect Bohr Model of Atom

Why this shape? Why the drop?

Blackbody Radiation A black body is an ideal system that absorbs all radiation incident on it The electromagnetic radiation emitted by a black body is called blackbody radiation c

Blackbody Approximation A good approximation of a black body is a small hole leading to the inside of a hollow object The hole acts as a perfect absorber The nature of the radiation leaving the cavity through the hole depends only on the temperature of the cavity

Blackbody Experiment Results The total power of the radiation emitted from the surface increases with temperature Stefan s law: P = σaet 4 P is the power and σ is the Stefan-Boltzmann constant: σ = 5.670 x 10-8 W / m 2. K 4 (For a blackbody, where e = 1) The peak of the wavelength distribution shifts to shorter wavelengths as the temperature increases Wien s displacement law λ max T = 2.898 x 10-3 m. K 5 4 k B 2 3 2π σ = 15c h = 5.67 10 8 W/m 2 K 4

Intensity of Blackbody Radiation I = P/A = σt 4 The intensity increases with increasing temperature The amount of radiation emitted increases with increasing temperature The area under the curve The peak wavelength decreases with increasing temperature

Rayleigh-Jeans Law An early classical attempt to explain blackbody radiation was the Rayleigh- Jeans law, based on the equipartition of energy: 2πck I (, ) BT λ T = 4 λ At long wavelengths, the law matched experimental results fairly well

Rayleigh-Jeans Law At short wavelengths, there was a major disagreement between the Rayleigh-Jeans law and experiment This mismatch became known as the ultraviolet catastrophe You would have infinite energy as the wavelength approaches zero

Max Planck: Father of Quantum Introduced the concept of quantum of action in 1900 In 1918 he was awarded the Nobel Prize for the discovery of the quantized nature of energy

Planck s Theory of Blackbody Radiation In 1900 Planck developed a theory of blackbody radiation that leads to an equation for the intensity of the radiation This equation is in complete agreement with experimental observations He assumed the cavity radiation came from atomic oscillations in the cavity walls Planck made two assumptions about the nature of the oscillators in the cavity walls

Planck s Two Assumptions The energy of an oscillator can have only certain discrete values E n = nhƒ This says the energy is quantized Each discrete energy value corresponds to a different quantum state

Strings are Quantized The possible frequencies and energy states of a wave on a string are quantized. f = v n 2 l

Continuous vs Discrete This is a continuous spectrum of colors: all colors are present. This is a discreet spectrum of colors: only a few are present.

Planck s Two Assumptions The energy of an oscillator can have only certain discrete values E n = nhƒ This says the energy is quantized Each discrete energy value corresponds to a different quantum state The oscillators emit or absorb energy when making a transition from one quantum state to another The entire energy difference between the initial and final states in the transition is emitted or absorbed as a single quantum of radiation

Why is quantum energy NEW & WEIRD????? In classical physics particles have continuous energy states.to say they have discrete energy states would mean that you can only drive at 10mph and 20mph but not at 15mph or at any speed in between 10 and 20 mph!

Energy-Level Diagram An energy-level diagram shows the quantized energy levels and allowed transitions Energy is on the vertical axis Horizontal lines represent the allowed energy levels The double-headed arrows indicate allowed transitions

More About Planck s Model The average energy of a wave is the average energy difference between levels of the oscillator, weighted according to the probability of the wave being emitted This weighting is described by the Boltzmann distribution law and gives the probability of a state being occupied as being proportional to B e EkT

Planck s Wavelength Distribution Function Planck generated a theoretical expression for the wavelength distribution I λ, T ( ) = λ 5 2πhc 2 ( hc λk 1) B e T h = 6.626 x 10-34 J. s h is a fundamental constant of nature

Planck s Model, Graphs

Intensity of Blackbody Radiation P40.61 The total power per unit area radiated by a black body at a temperature T is the area under the I(λ, T) versus λ curve, as shown in Figure 40.3. (a) Show that this power per unit area is 0 I 4 ( λ, T ) dλ = σt where I(λ, T) is given by Planck s radiation law and σ is a constant independent of T. This result is Stefan s law. To carry out the integration, you should make the change of variable x = hc/λkt and use the fact that 3 4 x dx π = 0 x e 1 15

Atomic Energy is quantized. It comes in chunks of Planck s constant, h. Max Planck NEVER liked the idea of quantized energy states. E = nhf, n= 0,1,2,3,... 34 h= 6.626x10 Js

The Photoelectric Effect Proof that Light is a Particle The Problem with Waves: Increasing the intensity of a low frequency light beam doesn t eject electrons. This didn t agree with wave picture of light which predicts that the energy of waves add so that if you increase the intensity of low frequency light (bright red light) eventually electrons would be ejected but they don t! There is a cut off frequency, below which no electrons will be ejected no matter how bright the beam! Also there is no time delay in the ejection of electrons as the waves build up!

Photoelectric Effect Apparatus When the tube is kept in the dark, the ammeter reads zero When plate E is illuminated by light having an appropriate wavelength, a current is detected by the ammeter The current arises from photoelectrons emitted from the negative plate and collected at the positive plate

Photoelectric Effect Apparatus

Photoelectric Effect Problem 1 Dependence of photoelectron kinetic energy on light intensity Classical Prediction Electrons should absorb energy continually from the electromagnetic waves As the light intensity incident on the metal is increased, the electrons should be ejected with more kinetic energy Experimental Result The maximum kinetic energy is independent of light intensity The maximum kinetic energy is proportional to the stopping potential (ΔV s )

Photoelectric Effect Problem 2 Time interval between incidence of light and ejection of photoelectrons Classical Prediction At low light intensities, a measurable time interval should pass between the instant the light is turned on and the time an electron is ejected from the metal This time interval is required for the electron to absorb the incident radiation before it acquires enough energy to escape from the metal Experimental Result Electrons are emitted almost instantaneously, even at very low light intensities

Photoelectric Effect Problem 3 Dependence of ejection of electrons on light frequency Classical Prediction Electrons should be ejected at any frequency as long as the light intensity is high enough Experimental Result No electrons are emitted if the incident light falls below some cutoff frequency, ƒ c The cutoff frequency is characteristic of the material being illuminated No electrons are ejected below the cutoff frequency regardless of intensity

Photoelectric Effect Problem 4 Dependence of photoelectron kinetic energy on light frequency Classical Prediction There should be no relationship between the frequency of the light and the electric kinetic energy The kinetic energy should be related to the intensity of the light Experimental Result The maximum kinetic energy of the photoelectrons increases with increasing light frequency

E = hf Light is quantized in chunks of planck s constant. Electrons will not be ejected in the Photoelectric Effect unless every photon has the right energy. One photon is completely absorbed by each electron ejected from the metal. As you increase the intensity of the beam, more electrons are ejected, but their energy stays the same. Photons

hf = KEmax + ϕ Work to eject (Work Function) Photon Energy Max KE of ejected electron

Photon Model Explanation of the Photoelectric Effect Dependence of photoelectron kinetic energy on light intensity K max is independent of light intensity K depends on the light frequency and the work function Time interval between incidence of light and ejection of the photoelectron Each photon can have enough energy to eject an electron immediately

Photon Model Explanation of the Photoelectric Effect Dependence of ejection of electrons on light frequency There is a failure to observe photoelectric effect below a certain cutoff frequency, which indicates the photon must have more energy than the work function in order to eject an electron Without enough energy, an electron cannot be ejected, regardless of the light intensity

Photon Model Explanation of the Photoelectric Effect Dependence of photoelectron kinetic energy on light frequency Since K max = hƒ φ As the frequency increases, the kinetic energy will increase Once the energy of the work function is exceeded There is a linear relationship between the kinetic energy and the frequency

The Photoelectric Effect What is the maximum velocity of electrons ejected from a material by 80nm photons, if they are bound to the material by 4.73eV? Ignore relatavistic effects. KMax = hf BE= hc BE λ ( )( ) 34 8 6.63 10 J s 3.00 10 m s 19 1.60 10 J = 9 ( 4.73 ev ) 80.0 10 m 1 ev K = 18 1.7295 10 J ( )( 2 1.729 10 18 J) 12 1 2 2KE 6 = mv v = = = 1.95 10 m s 31 2 m 9.11 10 kg 2 (SR: K = ( γ 1) mc )

Arthur Holly Compton 1892-1962 Director of the lab at the University of Chicago Discovered the Compton Effect, 1923 Shared the Nobel Prize in 1927

Compton Effect, Classical According to the classical theory, EM waves incident on electrons should: have radiation pressure that should cause the electrons to accelerate set the electrons oscillating Predictions

Compton Effect, Observations Compton s experiments showed that, at any given angle, only one frequency of radiation is observed

Compton Effect, Explaned The results could be explained by treating the photons as point-like particles having energy hƒ Assume the energy and momentum of the isolated system of the colliding photonelectron are conserved This scattering phenomena is known as the Compton effect

Compton Shift Equation The graphs show the scattered x-ray for various angles The shifted peak, λ is caused by the scattering of free electrons h λ' λ ( 1 cos ) o = θ mc e This is called the Compton shift equation

The photon transfers momentum, like a particle. p v = p v + p v h λ' λo = 1 cosθ mc e incident scattered electron ( ) p = h λ

Compton Problem 27. In a Compton scattering experiment, an x ray photon scatters through an angle of 17.4 from a free electron that is initially at rest. The electron recoils with a speed of 2180 km/s. Calculate (a) the wavelength of the incident photon and (b) the angle through which the electron scatters. h λ' λo = 1 cosθ mc e ( )

If photons can be particles, then why can t electrons be waves? p hf = E/ c= = c h λ debroglie Wavelength: λ = e h p

Electrons exist in quantized orbitals with energies given by multiples of Planck s constant. Light is emitted or absorbed when an electron makes a transition between energy levels. The energy of the photon is equal to the difference in the energy levels: E = nhf, n= 0,1,2,3,... = 34 h 6.626x10 Js E E E hf γ = i f =

Bohr s Model Explains Hydrogen Spectrum E E E hf γ = i f =

Bohr assumed that the electrons exist in stable orbits that they do not radiate as they orbit (accelerate!). This was a wild assumption with no explanation. E = nhf, n= 0,1,2,3,... = 34 h 6.626x10 Js E E E hf γ = i f =

1924: de Broglie Waves Electrons are STANDING WAVES in atomic orbitals. λ = h p 2π r n = n λ

De Broglie Wavelength λ = e h mv 34 h= 6.626x10 J s

Electron De Broglie Wavelength for electron v =.1c λ = h/ mv λ = 6.626x10 34 J s 31 7 (9.1x10 kg)(3x10 m / s) λ = 2.4x10 11 m

Particle Wave Duality What is a particle? What is a wave?

Particles: Rock Like Particles have mass (inertia). Particles are localized in space: Local Particles have extension and occupy space. Two particles can not occupy the same space. A particle that is moving has MOMENTUM. Momentum = mass x velocity Particles that are moving also have kinetic energy depending on the speed of the particle. Particles can have a continuous spectrum of speeds and energies. Energies are not quantized. Measuring the speed or position of a macroscopic particle does not disturb the particle.

result from periodic disturbance Characterized by 1 f = Τ periodor frequency (time for each bit to go through one cycle, up and down or to and fro.) wavelength (the distance the cycle repeats in a time freeze frame, distance peak to peak or trough to trough) wave speed (how fast is the energy transferred) v = λ f

Waves Transmit Energy Wave Energy is proportional to frequency: the faster he sends a pulse down the string, the more energy transmitted to the dog! Arrrrf!

Superposition Waves ADD in space. Simply add them point by point.

Interference Waves ADD: Constructive Interference. Waves SUBTRACT: Destructive Interference. In Phase Out of Phase

Interference: Two Spherical Sources

Superposition Traveling waves move through each other, interfere, and keep on moving!

Superposition Traveling waves move through each other, interfere, and keep on moving!

Standing Waves Produced by the superposition of two identical waves moving in opposite directions.

Standing Waves on a String Harmonics

Standing Waves Standing waves form in certain MODES based on the length of the string or tube or the shape of drum or wire. Not all frequencies are permitted!

Particles & Waves Spread Out in Space: NONLOCAL Superposition: Waves add in space and show interference. Do not have mass or Momentum Waves transmit energy. Bound waves have discreet energy states they are quantized. Localized in Space: LOCAL Have Mass & Momentum No Superposition: Two particles cannot occupy the same space at the same time! Particles have energy. Particles can have any energy.

Particle-Wave: Light A gamma ray photon has a momentum of 8.00x10-21 kg m/s. What is its wavelength? What is its energy in Mev? h λ = = p 34 663. 10 J s 21 800. 10 kg m s = 829. 10 14 m E ( 21 )( 8 8.00 10 kg m s 3.00 10 m s) = pc = 1 MeV = = 1.60 10 J 12 2.40 10 J 15.0 MeV 13

Limits of Vision Electron Waves 11 λ e = 2.4x10 m

Electron Microscope Electron microscope picture of a fly. The resolving power of an optical lens depends on the wavelength of the light used. An electron-microscope exploits the wave-like properties of particles to reveal details that would be impossible to see with visible light.

Electron Microscope Salmonella Bacteria Stem Cells The fossilized shell of a microscopic ocean animal is magnified 392 times its actual size.

Lynda s De Broglie Wavelength λ = 6.626x10 34 J s (75 kg)2 m / s λ = 4.4x10 36 m Too small to notice or to interact with anything!

Particle Wave Duality

I E 2

If electron were hard bullets, there would be no interference pattern.

In reality, electrons do show an interference pattern, like light waves.

Electrons act like waves going through the slits but arrive at the detector like a particle.

Interference pattern builds one electron at a time. Electrons act like waves going through the slits but arrive at the detector like a particle.

Electron Diffraction

Wave Packet: Making Particles out of Waves h p = c= λ f λ p = hf / c Superposition of waves to make a defined wave packet. The more waves used of different frequencies, the more localized. However, the more frequencies used, the less the momentum is known.

Wave Packet Diffraction

Heisenberg Uncertainty Principle ΔΔ x p h/4π You make a wave packet by wave superposition and interference. The more waves you use, the more defined your packet and the more defined the position of the particle. However, the more waves you use of different frequencies (energy or momentum) to specify the position, the less you specify the momentum!

Double Slit for Electrons shows Wave Interference

Trying to see what slit an electron goes through destroys the interference pattern.

Feynman version of the Uncertainty Principle It is impossible to design an apparatus to determine which hole the electron passes through, that will not at the same time disturb the electrons enough to destroy the interference pattern. -Richard Feynman

Heisenberg Microscope Short wavelength of light must be used to find the electron. The shorter the wavelength, the more specified the location. But short wavelength means high energy. That energy is transferred to the electron in an unpredictable way and the motion (momentum) becomes uncertain. If you use long wavelength light, the motion is not disturbed but the position is uncertain. Δx ~ λ Δ p= h/ λ ΔxΔ p> h

Electron in a Box A wave packet in a square well (an electron in a box) changing with time.

The possible wavelengths for an electron in a box of length L. Electron in a Box Δx~ L Δ p~ p~ h/ λ = h/ L If you squeeze the walls to decrease Δx, you increase Δp! ΔΔ x p~ L h/ L> h

Electron Waves leads to Quantum Theory Waves: De Broglie: 2 L λ =, n = 1,2,3... n n h λ = E p 2 1 p = mv 2 = 2 2m Combine: E n 2 2 hn = Energy is Quantized! 8mL 2