The experimental basis of quantum theory

The experimental basis of quantum theory

Preliminary remarks New theories do not appear from nowhere, they are usually based on (unexplained) experimental results. People have to be ready for it, e.g. be able to interpret what they see as a new phenomenon and not an experimental flow in their setup. This also means being aware of new developments in physics Louis Pasteur (France, 1822 1895) In the fields of observation, chance only favors the prepared mind (1854)

Part I: Blackbody Radiation Max Planck 1858-1947 Nobel Prize in Physics 1918

Thermal Radiation Known since centuries that when a material is heated, it radiates heat and its color depends on its temperature Example: heating elements of a stove: Dark red: 550ºC Bright red: 700ºC Then: orange, yellow and finally white (really hot!) The emission spectrum depends on the material Theoretical description: simplifications necessary à Blackbody Thermal images taken before and after the zombie apocalypse

A material is constantly exchanging heat with its surrounding (to remain at a constant temperature): It absorbs and emits radiations Problem: it can reflect incoming radiations, which makes a theoretical description more difficult (depends on the environment) A blackbody is a perfect absorber: Incoming radiations is totally absorbed and none is reflected Blackbody Blackbody = a cavity, such as a metal box with a small hole drilled into it. Incoming radiations entering the hole keep bouncing around inside the box with a negligible chance of escaping again through the hole à Absorbed. The hole is the perfect absorber, e.g. the blackbody Radiation emission does not depend on the material the box is made of à Universal in nature

Blackbody radiation

Wien s displacement law The intensity (λ, T) is the total power radiated per unit area per unit wavelength at a given temperature Wien s displacement law: The maximum of the distribution shifts to smaller wavelengths as the temperature is increased. Originally an empirical formula Wilhem Wien Nobel Prize (Physics) 1911 Visible light: 400 700 nm Ultra-violet: <400 nm Infrared: >700 nm

Exercise - blackbody Dominant color of a blackbody at: T=4000ºC l = 678 nm RED T=5000ºC l = 549 nm GREEN T=6000ºC l = 461 nm BLUE

Stefan-Boltzmann Law The total power radiated per unit area increases with the temperature: This is known as the Stefan-Boltzmann law, with the constant σ experimentally measured to be 5.6705 10 8 W / (m 2 K 4 ). The emissivity є (є = 1 for an idealized blackbody) is simply the ratio of the emissive power of an object to that of an ideal blackbody and is always less than 1.

Understanding the blackbody spectrum Attempts to fit the low and high wavelength part of the spectrum Using classical theory of electromagnetism and thermodynamics, Lord Rayleigh comes up with: Rayleigh-Jeans formula Major flaw at short wavelength ( Ultraviolet catastrophe ) Describing the blackbody emission spectra: one of the outstanding problems at the beginning of the 20 th century

Two catastrophes? Classical physics: Emission spectrum: a superposition of electromagnetic waves of different frequencies Frequencies allowed: standing waves inside the cavity Equipartition of the energy: Every standing wave carries kt of energy Flaw: when l à 0, the number of standing waves à, leading to E à [Ultraviolet Catastrophe] Failure of classical theories: The work of Rayleigh-Jeans was considered as state-of-the-art, using well tested theories, which were in very good agreement with experimental results in many other circumstances. Need for a new theory

Max Planck and the blackbody problem Max Planck 1858-1947 Expert in thermodynamics and statistical mechanics Around 1900: Proposes first an empirical formula (based on real physics) to reproduce both the high and low wavelength parts of the emission spectrum à Remarkable agreement with experimental results Then, works on a theoretical basis of the formula

Planck s radiation law Planck assumed that the radiation in the cavity was emitted (and absorbed) by some sort of oscillators contained in the walls. He used Boltzman s statistical methods to arrive at the following formula: Planck s radiation law Planck made two modifications to the classical theory: The oscillators (of electromagnetic origin) can only have certain discrete energies determined by E n = nhn, where n is an integer, n is the frequency, and h is called Planck s constant. h = 6.6261 10 34 J s. The oscillators can absorb or emit energy in discrete multiples of the fundamental quantum of energy given by DE = hn

Quantization! Blackbody emission spectrum explained by introducing quantization of energy transfers, resolves the ultraviolet catastrophe Low wavelength ßà High frequency (n = c/l) At small l, the energy E=hn needed to fill up the oscillator states increases. Their probability to be occupied decreases rapidly, e.g. faster than the rate found in the Rayleigh-Jeans formula: no ultraviolet catastrophe. Very disputed Planck himself looked for a few years in ways to get hà0 without success.

Quantum theory needed! Planck s radiation law Power radiated at a given frequency for a given blackbody temperature Power radiated at a given wavelength for a given blackbody temperature: Planck s radiation law

PHGN324: Why is blackbody radiation relevant to astronomy & astrophysics? A blackbody is a perfect absorber: incoming radiations is totally absorbed and none is reflected. The Sun (and any other stars) can be approximated to a Black Body: Almost a perfect absorber (Near) thermal equilibrium At the top of the atmosphere

PHGN324: Star color / temperature / Luminosity CLASSIFICATION: TEMPERATURE (K): Luminosity: where Hertzsprung-Russell (H-R) diagram Luminosity vs temperature L = 4π R 2 σt 4 R the radius of the star T the temperature of the star s the Stefan-Boltzmann constant (s=5.67x10-8 W.m -2.K -4 )

PHGN324: The Cosmic Microwave Background (CMB) Cosmic Microwave Background (CMB) CMB anisotropy 16 µk, DT/T=5 x 10-6 The CMB suggests that, at some point, the Universe was extremely dense and hot, and filled with radiation in THERMAL EQUILIBRIUM = Blackbody. After the time of last scattering (T~3000K - when the universe becomes transparent to radiation), the radiation cools off (redshift) due to the expansion of the Universe (now: T~2.728K)

Exercise Black hole temperature! A black hole may well be the perfect absorber. Famous astrophysicist Stephen Hawking suggests that a black hole can radiate energy with a thermal spectrum due to quantum effects (Hawking radiation). Lets consider a black hole as a sphere with a radius of 30km radiating 8.8x10-31 W of such thermal radiation. What would be the temperature of this black hole (in K)? [Hint: remember that the power is radiated from the surface of the black hole].

Part II: X-ray, Electrons and Line Spectra

Cathode rays Known since 1800 s Produced by a large difference of potential in an evacuated tube.

A brief history of cathode rays (and vacuum) 1833: Michael Faraday (1791-1867) Study of electrical discharge of gas Notices that: rarefaction of the air wonderfully favors the glow phenomena à Poor vacuum 1858: Julius Plücker (1801-1868) Seems to observe a deflection of the rays, when approaching a magnet Bright green glow around the cathode? à Still poor vacuum 1869 Johann Hittorf (1824-1914) Reports that the glowing originates from the cathode Observes sharp shadows, the rays are traveling in straight lines à Better vacuum 1879 William Crookes reaches 40x10-3 mm Hg

Approaching the 20 th century The debate heats up: 1892: Heinrich Hertz claims experimental evidences that cathode rays cannot be particles 1895: Jean-Baptiste Perrin shows that cathode rays are negatively charged particles Cathode rays: Waves or Particles? In 1895, no one knows what cathode rays really are, but some of their properties were measured. So, the studies go on

The discovery of X-rays (I) Wilhem Röntgen (1845-1923) on the evening of Nov 8, 1895 While working on cathode rays, notices that more penetrating rays are emitted from the interaction of the cathode rays with the tube, fluorescence on a nearby screen Measures the transmission of the rays through various materials. They appear to be transparent to the rays. Finally put his hand in front of rays and sees his bones on the screen!

The discovery of X-rays (II) Röntgen does not say anything to anybody. Astonished by his discovery, needs to check, double-check, triplecheck for about 7 weeks! Properties of the X-rays: More penetrating than cathode rays Not deviated by electric/magnetic fields Immediate medical applications! Finally, on Dec 28, 1895, he submits a preliminary paper: On a new kind of rays which is distributed in January 1896 By the end of 1896, more than a thousand papers were written on the subject of X- rays. 1902: Röntgen receives the first Nobel Prize in Physics

Nature of the cathode rays: the discovery of the electron J.J.Thomson 1856-1940 (Nobel Prize 1906) Manages to show conclusively that cathode rays are deflected by electric and magnetic fields; do not depend on the material the cathode or the anode is made of. Have a negative charge, cathode rays are particles Measures e/m ratio, close (about 35% off) to the present value of 1.76x10 11 C/kg à Better vacuum available after progress by Crookes

An electron moving through the electric field is accelerated by a force: F + = ma + = qe = ee Electron angle of deflection: Measurement of the e/m ratio tan θ = v + v 4 = a +t v / = ee m The magnetic field deflects the electron against the electric field force. F = qe + qv B The magnetic field is adjusted until the net force is zero. l v / 7 E = v B v = E B = v / Charge-to-mass ratio: e m = v / 7 tan θ El = E tan θ B 7 l

Determination of the electron charge Robert Millikan 1868-1953 oil drop experiment (1907-1911) Used an electric field and gravity to suspend a charged oil drop. 8 F = 0 e E = m g Mass is determined from Stokes s relationship of the terminal velocity to the radius, and the density. m = 4 3 πra ρ Magnitude of the charge on the oil drop. e = mgd V Thousands of experiments showed that there is a basic quantized electron charge. e = 1.602 10 IJK C

The electron J.J. Thomson: M N = 1.76 10JJ C/kg R. Millikan: e = 1.6x10-19 C; Electron charge: -e à m e = 9.1x10-31 kg The e/m e ratio was much larger (x1000) than expected (based on results with the Hydrogen atom ). This is due to the fact that: à m(h + )=m(proton)~1836 x m e Conclusions: both the mass and the charge of the electron are quantified

Line spectra d sin θ = nλ With d: distance between slits. It is observed that chemical elements produce unique colors when burned (with a flame) or excited (with an electrical discharge) Diffraction creates a line spectrum pattern of light bands and dark areas on the screen. The line spectrum serves as a fingerprint of the gas that allows for unique identification of chemical elements and material composition.

Absorption vs Emission spectrum

The line spectra of stars (I) Absorption spectrum of stars: Inner, dense layers of the star produce a continuous (blackbody) spectrum Cooler surface layers absorb light at specific wavelengths / frequencies

The line spectra of stars (II): assessing how old is a star Metal-poor star (very old star) Metal-rich star (relatively young star) The Sun is a metal-rich star

Balmer series In 1885, Johann Balmer (a swiss schoolteacher) finds an empirical formula for wavelength of the visible hydrogen line spectra in nm: λ = 364.56 k7 k 7 4 nm (where k = 3,4,5 ) à Underlying order/quantification not understood

As more spectral lines are discovered, a more general empirical equation appears: the Rydberg equation: 1 λ = R Y 1 n 7 1 k 7 R Y = 1.096776 10 [ m IJ Rydberg equation Rydberg constant (for Hydrogen)

Exercise Find the Balmer formula from the Rydberg equation Determine a formula for the Lyman series (n=1) and the Pashen series (n=3)

Conclusions so far. Early experiments point to a quantization of certain quantities Quantization appears in empirical formula as a way to describe Nature Still no theory can explain the observed behaviors Does an accurate description of Nature indeed require quantization?

Part III: The Photoelectric Effect

The photoelectric effect Photoelectric effect: electrons are emitted from any surface (especially clean metal surface) when light of a sufficiently high-frequency (e.g. short wavelength) shines on that surface

Typical experimental setup Incident light triggers the emission of (photo)electrons from the cathode Some of them travel toward the collector (anode) with an initial kinetic energy The applied voltage V either accelerates (if positive) or decelerates (if negative) the incoming electrons. The intensity I of the current measured by the ammeter as a function of the applied voltage V is a measurement of the photoelectron properties, and therefore a measurement of the properties of the photoelectric effect.

Measurements Work done mostly by Philipp Lenard (Nobel prize in Physics 1905) Known properties since about 1902 When applying a negative (retarding) potential, it is possible to stop all the incoming electrons: K max (e - ) = ev 0

Measurements

Measurement #1 The amount of photoelectrons is proportional to the incident light intensity (same frequency) The retarding potential does not depend on the incoming light intensity: K max is not a function of I Classical: K max should be a function of I

Measurement #2 The retarding potential depends on the frequency: Higher frequencies generates higher energy electrons Classical theory: cannot explain this either!

Work function Work function f: minimum extra energy that allows electrons to escape the material. ev (electron-volt: energy gained by an electron in an accelerating potential of 1V) New unit of energy: 1 ev = 1.6 x 10-19 J

Measurement #3 The smaller the work function f of the emitter material, the smaller is the threshold frequency f of the light that can eject photoelectrons No electrons are produced below this frequency whatever the intensity of the incident light. Classical theory: cannot explain this either!

Measurement #4 Photoelectric current is proportional to the light intensity the number of photoelectrons produced is proportional to the intensity of the incoming light Classical theory: Yes!

Takes Planck s ideas a step further Einstein s theory (1905) Suggests that the electromagnetic radiation field itself is quantized. Quanta of light carry a energy E=hn (photons) h: Planck s constant ; n: frequency of light Travels at the speed of light: ln = c Photoelectric effect: when a photon collides with an electron, it gives away all its energy (which is transferred to the electron as kinetic energy) Collision: Photon = Particle! Nobel Prize 1921 ( for the theory of the photoelectric effect and other significant contributions

The miracle year 1905 Theory of the Photoelectric Effect (p.132) Explanation of the Brownian Motion (p. 549) Special Theory of Relativity (p.891) Published in the same volume of Annalen der Physik, volume 17 (1905)

Einstein s theory Conservation of the energy: hn = f + K e - Energy of the incoming photon Kinetic Energy of the electron Kinetic Energy of the electron (non-relativistic): Work function (e.g. energy necessary for the electron to escape the material) Experimentally: K e - max = (1/2)m e -v max 2 = ev 0 Retarding Potential

Einstein s theory From hυ = φ + K c M`ab hυ φ = ev / With One gets φ = hυ / h υ υ / = ev / Frequency: f or n

Exercise What is the threshold frequency n 0 for the photoelectric effect on lithium (f=2.93ev)? What is the stopping potential if the wavelength of the incident light is 400nm?

Part IV: Compton scattering

The Compton effect Another compelling proof of the particle nature of light: The scattered photon is measured to have a longer wavelength (lower frequency) than the original photon à Cannot be explained by classical e.m.

Conservation of Energy: Conservation of Momentum: Combining the two conservation laws: With υ = d e and λ = λg λ, one finds: Analysis (see derivation) E : hυ + m M c 7 = hυ g + m M c 7 7 + p M c 7 p 4 : hυ c = hυg c cos θ + p M cos φ p + : 0 = hυg c sin θ p M sin φ hυg c sin θ = p M sin φ p 7 4 + (p + ) 7 p 7 M = ( hυ c )7 +( hυg c )7 2( hυ c )(hυg ) cos θ c E 7 h υ υ g + m M c 7 7 = m M c 7 7 + ( op d )7 +( opq d )7 2( op m M c 7 υ υ g = hυυ g 1 cos θ λ = λ g λ = h m M c d )(opq d ) cos θ 1 cos θ

Analysis (f = υ, f g = υ g )

Experimental results

Exercise What is the energy of the photon resulting from the scattering of a 662-keV g-ray at 30º?

Pair production

Annihilation Matter Antimatter collision à release of pure energy