Stellar Astrophysics: The Interaction of Light and Matter The Photoelectric Effect Methods of electron emission Thermionic emission: Application of heat allows electrons to gain enough energy to escape Secondary emission: The electron gains enough energy by transfer from another high-speed particle that strikes the material from outside Field emission: A strong external electric field pulls the electron out of the material Photoelectric effect: Incident light (electromagnetic radiation) shining on the material transfers energy to the electrons, allowing them to escape 1
The Photoelectric Effect Electromagnetic radiation interacts with electrons within metals and gives the electrons increased kinetic energy Light can give electrons enough extra kinetic energy to allow them to escape The ejected electrons are called photoelectrons The minimum extra kinetic energy which allows escape is called the work function φ of the material Experimental Setup Minimum voltage for which I = 0 is called stopping potential V 0 2
Experimental Results The kinetic energies of the photoelectrons are independent of the light intensity The maximum kinetic energy of the photoelectrons, for a given emitting material, depends only on the frequency of the light The smaller the work function φ of the emitter material, the smaller is the threshold frequency of the light that can eject photoelectrons Experimental Results When the photoelectrons are produced, however, their number is proportional to the intensity of light The photoelectrons are emitted almost instantly following illumination of the photocathode, independent of the intensity of the light 3
Einstein s Theory Einstein suggested that the electromagnetic radiation field is quantized into particles called photons Each photon has the energy quantum Ephoton = h ν = Albert Einstein (1879-1955) hc λ where ν is the photon frequency and h is Planck s constant The photon travels at the speed of light in a vacuum, and its wavelength is given by λ = c ν Einstein s Theory Conservation of energy yields h ν = φ + Kelectron Explicitly, the energy of the most energetic electrons is Kmax = h ν φ The retarding potentials measured in the photoelectric effect are the opposing potentials needed to stop the most energetic electrons e V0 = Kmax 4
Quantum Interpretation The kinetic energy of the electron does not depend on the light intensity at all, but only on the light frequency and the work function of the material K max = e V 0 = h ν φ Einstein in 1905 predicted that the stopping potential was linearly proportional to the light frequency, with a slope h, the same constant found by Planck The Compton Effect A photon impinging on a material will scatter from an atomic electron For high photon energies, we can neglect the binding energy and treat the collision as an elastic collision between the photon and electron For the photon energy we can write Arthur Compton (1892-1962) E photon = h ν = h c λ = p c 5
The Compton Effect Scattering of photons off loosely bound atomic electron Conservation laws h ν i + m e c 2 = h ν f + p 2 e c 2 + (m e c 2 ) 2 h / λ i = h / λ f cos θ + p e cos φ h / λ f sin θ = p e sin φ The Compton Effect The scattering off the electron yields a change in wavelength of the scattered photon which is known as the Compton Effect Δ λ = λ f λ h I = ( 1 cos θ ) m e c = λ C ( 1 cos θ ) with Compton wavelength λ C = 0.00243 nm The radiation pressure discussed previously is caused by the Compton Effect 6
Line Spectra Chemical elements were observed to produce unique wavelengths of light when burned or excited in an electrical discharge Emitted light is passed through a diffraction grating with thousands of lines per cm and diffracted according to its wavelength λ by the equation where d is the distance between grating lines d sinθ = n λ with n = 0, 1, 2, Balmer Series In 1885, Johann Balmer found an empirical formula for wavelengths of the visible hydrogen line spectrum 1 λ 1 = R H 4 1 n 2 Johann Balmer (1825-1898) where n = 3,4,5 and R H = 1.097 10 7 m -1 Rydberg constant 7
De Broglie Waves Prince Louis V. de Broglie suggested that massive particles should have wave properties similar to electromagnetic radiation He was guided by the relations for photons h ν = p c = p λ ν Similarly for photons, the wavelength and momentum are related λ = h / p The wavelength of the particle, deduced from its momentum, is called the de Broglie wavelength Louis de Broglie (1892-1987) The Classical Atomic Model Let s consider atoms as a planetary model The force of attraction on the electron by the nucleus and Newton s second law give 1 4 π ε 0 e 2 r 2 = µ υ 2 r where υ is the tangential velocity of the electron and the reduced mass m µ = e m p m e + m p Can estimate υ, taking r = 5 10-11 m υ 2 10 6 m/s < 0.01 c 8
The Classical Atomic Model The total energy of the atom can be written as the sum of kinetic and potential energy E = K + U = 1 e 2 1 8 π ε 0 r 4 π ε 0 e 2 r = 1 8 π ε 0 e 2 r Problems with the Planetary Model From classical E&M, an accelerated electric charge radiates energy (electromagnetic radiation) So the total energy must decrease radius r must decrease Electron crashes into the nucleus in ~ 10-9 s!? 9
The Bohr Model of the Hydrogen Atom Bohr s assumptions Stationary states in which orbiting electrons do not radiate energy exist in atoms Emission/absorption of energy occurs along with atomic transition between two stationary states Δ E = E 1 - E 2 = h ν Classical laws of physics do not apply to transitions between stationary states The angular momentum of the atom in a stationary state is a multiple of h/2π L = n h 2 π n ћ n = principle quantum number Niels Bohr (1885-1962) Bohr s Quantization Condition The electron is a standing wave in an orbit around the proton This standing wave will have nodes and be an integral number of wavelengths h 2 π r = n λ = n p The angular momentum becomes L = r p = n h 2 π = n ћ 10
From Bohr Radius e 2 L = n h 2π n ћ and K = 8πε = 0 r µ υ 2 2 The radius of the hydrogen atom for stationary states is 4 π ε r n = 0 n 2 ћ 2 = n 2 a 0 µ e 2 quantized orbit (position) where the Bohr radius is a 0 = 4πε 0ћ 2 µ e 2 = 0.53x 10 10 m The smallest radius of the hydrogen atom occurs when n = 1 This radius gives its lowest energy state (called the ground state of the atom) The Hydrogen Atom The energies of the stationary states (with E 0 = 13.6 ev) e 2 e 2 E E n = = = 0 8 π ε 0 r n 8 π ε a 0 0 n2 n 2 Emission of light occurs when the atom in an excited state decays to a lower energy state (n u n l ) h ν = E h E l where ν is the frequency of the photon 1 λ = ν = E h E l n 2 h c = R 1 c 1 H l n h 2 and R H the Rydberg constant 11
Transitions in the Hydrogen Atom Lyman series (invisible) The atom exists in the excited state for a short time before making a transition to a lower energy state with emission of a photon. At ordinary temperatures, almost all hydrogen atoms exist in n = 1 state. Absorption therefore gives the Lyman series. Balmer series (visible) When sunlight passes through the atmosphere, hydrogen atoms in water vapor absorb wavelengths of the Balmer series giving dark lines in the absorption spectrum. Limitations of the Bohr Model The Bohr model was a great step for the new quantum theory, but it had its limitations. Works only for single-electron atoms Could not account for the intensities or the fine structure of the spectral lines Could not explain the binding of atoms into molecules 12
Wave Motion de Broglie matter waves led ultimately to a wave equation describing particle motion. First, let s review the physics of waves. The displacement of a wave is This is a solution to the wave equation Define the wave number k and the angular frequency ω as: and The wave function is now Ψ(x, t) = A sin (kx - wt) Wave Properties The phase velocity is the velocity of a point on the wave that has a given phase (for example, the crest) and is given by A phase constant φ shifts the wave. 13
Principle of Superposition When two or more waves traverse the same region, they act independently of each other Combining two waves yields Ψ(x,t) = Ψ 1 (x,t) + Ψ 2 (x,t) = 2Acos Δk 2 x Δω 2 t cos(k x ω t) av av The combined wave oscillates within an envelope that denotes the maximum displacement of the combined waves. By combining many waves with different amplitudes and frequencies, a pulse, or wave packet, can be formed which moves with a group velocity u gr = Δω / Δk. Fourier Series The wave packet may be expressed in terms of its component waves via a Fourier series Summing an infinite number of waves yields a Fourier integral 14
Wave Packet Envelope Superposition of two waves yields a wave number and angular frequency of the wave packet envelope The ranges of wave numbers and angular frequencies that produce the wave packet have the following relations A Gaussian wave packet has similar relations Δx = 1 2 λ = 1 2π env ΔkΔx = 2π 2 Δk /2 Δt = 1 2 T = 1 2π env ΔωΔt = 2π 2 Δω /2 The localization of the wave packet over a small region to describe a particle requires a large range of wave numbers. Conversely, a small range of wave numbers cannot produce a wave packet localized within a small distance. Gaussian Function A Gaussian wave packet is of the form: The group velocity is 15
Electron Double-Slit Experiment Claus Jönsson of Tübingen, succeeded in 1961 in showing double-slit interference effects for electrons by constructing very narrow slits and using relatively large distances between the slits and the observation screen This experiment demonstrated that precisely the same behavior occurs for both light (waves) and electrons (particles) Which Slit? To determine which slit the electron went through: We set up a light shining on the double slit and use a powerful microscope to look at the region. After the electron passes through one of the slits, light bounces off the electron; we observe the reflected light, so we know which slit the electron came through. Use a subscript ph to denote variables for light (photon). Therefore the momentum of the photon is p h ph = > The momentum of the electrons will be on the order of. The difficulty is that the momentum of the photons used to determine which slit the electron went through is sufficiently great to strongly modify the momentum of the electron itself, thus changing the direction of the electron! The attempt to identify which slit the electron is passing through will in itself destroy the interference pattern. λ ph p h el = ~ λ el h d h d 16
Wave Particle Duality Solution The solution to wave particle duality is given by the following principle Bohr s principle of complementarity: It is not possible to describe physical observables simultaneously in terms of both particles and waves Physical observables are those quantities such as position, velocity, momentum, and energy that can be experimentally measured In any given instance we must use either the particle description or the wave description Uncertainty Principle It is impossible to measure simultaneously, with no uncertainty, the precise values of k and x for the same particle. The wave number k may be rewritten as k = 2π λ = 2π h / p = p 2π h = p ћ For the case of a Gaussian wave packet we have ΔkΔx = Δp ћ Δx = 1 2 Thus for a single particle we have Heisenberg s uncertainty principle Δp x Δx ћ 2 The Gaussian packet turns out to give the minimum product of uncertainties (any other shape gives a larger product) 17