Guess now. It has been found experimentally that: (a) light behaves as a wave; (b) light behaves as a particle; (c) electrons behave as particles; (d) electrons behave as waves; (e) all of the above are true; (f) none of the above are true.
Radiation intensity is proportional to T 4 Materials glow white/yellow at 2000K.
The Wavelengths of the peaks follow λ p T = 2.90 10 3 This is Wien s law. Draw a graph of T against λ. What type of curve is it?
Example: Sun temperature If λ p =500 nm for the Sun, what is the temperature T = 2.90x10 3 500x10 9 = 6000K
Example: Star colour The surface temperature of a star is 32,500K. What colour would the star appear? Answer: λ p = 2.x10 3 32, 500 = 89.2nm This is the UV part of the spectrum. In the visible part the curve will be descending, therefore, the shortest visible wavelength will be the strongest. The star will, therefore, appear blue.
Planck s formula I = 2πhe2 λ 5 e hc λkt 1 Planck s assumption to justify the formula was that E = nhf, where n is an integer. n is called the quantum number. Planck suggested that energy of molecular vibration is only a multiple of a base energy value. Called Planck s quantum hypothesis.
Example: Photon energy. Calculate the energy of a photon of blue light λ = 450 nm (in air). Approach E = hf c = f λ E = hc λ Answer or 6.63x10 34 3.0x10 8 4.5 10 7 = 4.4 10 19 J 4.4 10 19 = 2.8eV 1.6 10 19
Example: Photons from a lightbulb. Esitmate how many visible light photons a 100W light-bulb emits per second. (Assume energy efficiency is 3% i.e 97% energy is lost as heat). Approach λ = 500nm E = hc λ Answer 3W = 3Js 1 n = 3 hf = 3 = nhc λ 3 500 10 9 6.63 10 34 3.0 10 8 = 8 1018
Exercise: Compare a light source of 1000nm (IR) with a light source of 100nm (UV). E = hf = hc λ = hc 1000 10 9 For λ = 1000nm n = 6.63 10 34 3.0 10 8 1000 1.6 10 38 12.43 = = 12.4 109 1000 10 12 λ = 100nm n = 12.4 10 10 which 10 times more electrons.
Photoeclectric effect Video: A simple gold leaf electroscope. Video: Lonnie s lab. UC Berkeley
What is the kinetic energy and the speed of an electron ejected from a sodium surface who s work function is W 0 = 2.28eV when (a) λ = 410nm (b) λ = 550nm Approach: Find the energy of the photons. E = hf = hc λ If E > W 0 we get electrons. The maximum kinetic energy is E W 0
Answer m = 9.11 10 31 kg. E = 1 2 mv 2 E = hc = 3.03 ev λ 2E E W 0 = 3.03 2.28 v = m 2 0.75 1.6 10 19 = 9.11 10 31 = 0.513 10 6 = 5.1 10 5 ms 1 (b) No electrons released. Note: If v is close to 0.1c we have to look at the relativistic equation.
Example 2 What is the lowest frequency and the longest wavelength needed to emit electrons from sodium? An exercise left for the student.
Energy, mass and momentum We know E = hf Photons travel at the speed of light. Need a relativistic formulae. (p mv for a photon) mv p = 1 v 2 c 2 Since v = c for a photon, p becomes infinite which is obviously wrong. Relativistic equation gives E 2 = p 2 c 2 + m 2 c 4 Setting m = 0 we get E 2 = p 2 c 2 p = E c Since E = hf for a photon, p = hf c = h λ
Example. The 10 19 photons emitted per sec from the 100W lightbulb are focused on a piece of black paper and absorbed. (a) Calculate the momentum of one photon. (b) Estimate the force of these photons on the paper. Approach: p = h. The momentum changes from p to zero. Use λ Newton s second law, F = p t. Answer p = h λ = 6.63. If N = 10 19 photons, F = 10 8 N.
Sun If the number of photons is large, n 10 8 N can be a considerable force! Solar sail is an example.
Photosynthesis Chlorophyll in plants captures the energy of sunlight to change CO 2 to useful carbohydrate. 9 photons are needed to transform one molecule of CO 2 to carbohydrate and O 2. Assuming λ = 670 nm (chlorophyll absorbs most strongly in the range 650nm - 700nm), how efficient is the photosynthetic process? The reverse chemical reaction releases an energy of 4.9eV/molecule of CO 2. Approach Efficiency is the minimum energy required (4.9eV) divided by the actual energy absorbed ( 9 the energy of one photon). Answer: hf = hc λ. 9 hcλ = 2.7 10 18 J. = 17eV. 4.9 17 = 29% efficient.
Compton effect. AH Compton (1892-1962) λ = λ + h (1 cosφ) m e c λ λ is the Compton shift. Used to measure bone density for osteoperosis.
Photon interactions - Summary We get: 1. Photoelectric effect. Knocks an electron out of an atom. Photon disappears. 2. Excited state of atom (see later). Photon disappears. 3. Scattered electron and photon. Compton effect. 4. Pair production. Creates matter (e.g. e + and e ). Photon disappears.
Wave nature of electron. Louis de Broglie (1892-1987) He proposed that the wavelength of a material particle would be related to its momentum (in same way as for a photon). p = h λ which implies λ = h p Note that this is valid classically, p = mv for (v << c) and mv relativistically p = 1 v. 2 c 2 λ = h is called the de Broglie wavelength of a particle. p
Example. Wavelength of a ball. Calculate the de Broglie wavelength of a 0.20 kg ball moving with speed of 15ms 1. Approach Use λ = h p. Answer λ = 2.2 10 34 m
Example What is the wavelength of an electron that has been accelerated through a potential of 100V. Approach Kinetic energy = 1 2 mv 2 = ev, λ = h mv Answer λ = 1.2 10 10 m = 0.12nm.
Electron diffraction Assume electrons strike perpendicular to the surface of a solid and their energy is low. K = 100eV. If the smallest angle at which diffraction maximum occurs is 24, what is the separation between the atoms on the surface.
Electron diffraction Approach Treat electrons as waves. Constructive interference occurs when the difference in path = λ, that is, a sin θ = λ K = p2 = h2 2m e 2m e λ 2 λ = a = λ sin θ = 0.123 = 0.30 nm. sin 24 h 2m e K = 0.123nm.
No-one has actually seen an electron but... they are good for looking at small things. The ebola virus attacks a cell. Transmission electron microscope (x 50,000). Scanning electron microscope (x35,000)
Transmission electron microscope (x 50,000). Scanning electron microscope (x35,000)
Quantum theory and models of the atom Spectra I Excited gasses in a discharge tube produce discrete spectra.
The emission spectra is a characteristic of the material and can serve as a finger print. Also if a continuous spectrum passes through a rarefied gas, dark lines appear where the emission lines occurred. This is called an absorption spectrum.
Hydrogen is the simplest atom. One electron. Simplest spectrum. J.J. Balmer (1825-1898) showed that the 4 lines in the visible spectrum have wavelengths that fit 1 λ = R( 1 2 2 1 n 2 ) for n = 3, 4,... R is the Rydberg constant. R = 1.0974 10 7 m 1 Balmer lines extend into the UV region ending at λ = 365nm. Later experiments found that if 1 1 2, 1 3 2, 1 4 2. 1 was replaced with 22
In general 1 λ = R( 1 n 2 1 n 2 ) n = 1 for Lyman series. n = 2 for Balmer series. n = 3 for Paschen series.
The Bohr model Electrons in orbits. Angular momentum = L = mvr n = nh for n = 1, 2, 3,... 2π F = 1 (Ze)e 4πɛ 0 r 2 Coulomb s law. Newton s second law gives F = ma. The acceleration of the particle is a = v 2 1 4πɛ 0 Ze 2 r n 2 = mv 2 r n r n