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Early Quantum Mechanics Chapter 27

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Three Major Discoveries Light is both a wave and a particle Electron Orbitals are Quantized The electron (and all matter) is both a wave and a particle Planck Einstein Compton BohrDe Broglie (Later Schrodinger/Hei senberg)

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Wein’s Law Treat’s light solely as a wave Hot “blackbodies” radiate EM The hotter the object, the shorter the peak wavelength Sun (~6000 K) emits in blue and UV A 3000 K object emits in IR 2.90 X mK = peak T

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Wein’s Law: Ex 1 Estimate the temperature of the sun. The Suns light peak at about 500 nm (blue) 500 nm = 500 X m T = 2.90 X mK peak T = 2.90 X mK = 6000 K 500 X m

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Wein’s Law: Ex 2 Suppose a star has a surface temperature of about 3500 K. Estimate the wavelength of light produced X mK = peak T peak = 2.90 X mK T peak = 2.90 X mK = 8.29 X 10-7 m = 830 nm 3500 K

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Planck’s Quantum Hypothesis Energy of any atomic or molecular vibration is a whole number Photon – the light particle Photons emitted come in “packets” E = hf h = X J s(Planck’s constant)

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Photons: Ex 1 Calculate the energy of a photon of wavelength 600 nm. 600 nm X 1 X m = 6 X10 -7 m 1 nm c = f f = c/ = (3X10 8 m/s)/(6 X10 -7 m) = 5 X s -1 E = hf E = (6.626 X J s)(5 X s -1 ) = 3.3 X J

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Photons: Ex 1a Convert your answer from the previous problem to electron Volts (1 eV = 1.6 X J)

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Photons: Ex 2 Calculate the energy of a photon of wavelength 450 nm (blue light).

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Photons: Ex 2a Convert your answer from the previous problem to electron Volts (1 eV = 1.6 X J)

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Photons: Ex 3 Estimate the number of photons emitted by a 100 Watt lightbulb per second. Assume each photon has a wavelength of 500 nm. 500 nm X 1 X m = 5 X10 -7 m 1 nm c = f f = c/ = (3X10 8 m/s)/(5 X10 -7 m) = 6 X s -1

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100 Watts = 100 J/s (we are looking at 1 second) E = nhf n = E/hf n = 100 J (6.626 X J s)(6 X s -1 ) n = 2.5 X 10 20

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Photoelectric Effect (Einstein) When light shines on a metal, electrons are emitted Can detect a current from the electrons Used in light meter, scanners, digital cameras (photodiodes rather than tubes)

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Three Key Points 1.Below a certain frequency, no electrons are emitted 2.Greater intensity light produces more electrons 3.Greater Frequency light produces no more electrons, but the come off with greater speed

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Low FrequencyHigh Frequency Not enough energy to eject electron Can eject electron Energy of photon is greater than W (ionization energy

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2.More intensity –More photons –More electrons ejected with same KE

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3.Greater Frequency –No more electrons ejected –Electrons come off with greater speed (KE)

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hf = KE + W hf = energy of the photon KE =Maximum KE of the emitted electron W = Work function to eject electron

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MetalWork Function (eV) Na2.46 Al4.08 Cu4.70 Zn4.31 Ag4.73 Pt6.35 Pb4.14 Fe4.50

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hf = KE + W: Ex 1 What is the maximum kinetic energy of an electron emitted from a sodium atom whose work function (W o ) is 2.28 eV when illuminated with 410 nm light? 410 nm = 410 X m or 4.10 X m 2.28 eV X 1.60 X J=3.65 X J 1 eV

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c = f f = c/ f = c/(4.10 X m) = 7.32 X s -1 hf = KE + W KE = hf – W KE = (6.626 X J s)(7.32 X s -1 ) X J KE = 1.20 X J or 0.75 eV

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hf = KE + W: Ex 2 What is the maximum kinetic energy of an electron emitted from a sodium atom whose work function (W o ) is 2.28 eV when illuminated with 550 nm light? ANS: 2.25 eV

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hf = KE + W: Ex 3 What is the maximum wavelength of light (cutoff wavelength) that will eject an electron from an Aluminum sample? Aluminum’s work function (W o ) is 4.08 eV? 4.08 eV X 1.60 X J=6.53 X J 1 eV

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hf = KE + W hf = 0 + W(looking for bare minimum) c = f f = c/ hf = W hc = W = hc = (6.626 X J s)(3.0 X 10 8 m/s) W 6.53 X J = 3.04 X m = 304 nm (UV)

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Photon/Matter Interactions 1.Electron excitation (photon disappears) 2.Ionization/photoelectric effect (photon disappears) 3.Scattering by nucleus or electron 4.Pair production (photon disappears)

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Electron Excitation Photon is absorbed (disappears) Electron jumps to an excited state Ionization/Photoelectric Effect Photon is absorbed (disappears) Electron is propelled out of the atom

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Scattering Photon collides with a nucleus or electron Photon loses some energy Speed does not change, but the wavelength increases Pair Production Photon closely approaches a nucleus Photon disappears An electron and positron are created.

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3. Scattering: Compton Effect Electrons and nuclie can scatter photons Scattered photon is at a lower frequency than incident photon Some of the energy is transferred to the electron or nucleus

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’ = + h (1 – cos ) m o c ’ = wavelength of scattered photon = wavelength of incident photon m o = rest mass of particle = angle of incidence

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Compton Effect: Ex 1 X-rays of wavelength nm are scatterd from a block of carbon. What will be the wavelength of the X-rays scattered at 0 o ? ’ = + h (1 – cos ) m o c ’ = 140 X10 -9 m + (6.626 X J s)(1 – cos ) (9.11 X kg)(3 X 10 8 m/s) ’ = 140 X10 -9 m + 0 ’ = 140 nm

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Compton Effect: Ex 2 What will be the wavelength of the X-rays scattered at 90 o ? ’ = + h (1 – cos ) m o c ’ = 140 X10 -9 m + (6.626 X J s)(1 – cos 9 ) (9.11 X kg)(3 X 10 8 m/s) ’ = 140 X10 -9 m X m ’ = 142 nm

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Compton Effect: Ex 3 What will be the wavelength of the X-rays scattered at 180 o ? ’ = + h (1 – cos ) m o c ’ = 140 X10 -9 m + (6.626 X J s)(1 – cos 180) (9.11 X kg)(3 X 10 8 m/s) ’ = 140 X10 -9 m X m ’ = 145 nm (straight back)

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4. Pair Production Photon Disappears e - and e + are produced They have opposite direction (law of conservation of momentum) When e - and e + collide they annihilate each other a new photon appears

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Principle of Complimentarity Neils Bohr Any experiment can only observe light’s wave or particle properties, not both Different “faces” that light shows WaveParticle PrismBlackbody Radiation Photoelectric effect

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The Discovery of the Electron (Thomson) Cathode Ray Tube Charged particles produced (affected by magnetic field)

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Concluded that atom must have positive and negative parts Electron – negative part of the atom Only knew the e/m ratio Plum Pudding Model

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Charge and Mass of the Electron (Millikan) Oil drop experiment Determines charge on electron (uses electric field to counteract gravity) Quantized e = X C m = 9.11 X kg

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The Nucleus (Rutherford) Gold Foil Experiment Discovers nucleus (disproves Plum Pudding Model) Planetary Model

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Wave Nature of Matter Everything has both wave and particle properties DeBroglie Wavelength E 2 = p 2 c 2 + m 2 c 4 (consider a photon) E 2 = p 2 c 2 (photon has no mass) E = pc E = hf hf = pc c = f

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hf = p f p = mv (for a particle) hf = mv f h = mv = h mv Everything has a wavelength Diffraction pattern of electrons scattered off aluminum foil

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DeBroglie Wavelength: Ex 1 Calculate the wavelength of a baseball of mass 0.20 kg moving at 15 m/s = h mv = (6.626 X J s) (0.20 kg)(15 m/s) = 2.2 X m

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DeBroglie Wavelength: Ex 2 Calculate the wavelength of an electron moving at 2.2 X 10 6 m/s = h mv = (6.626 X J s) (9.11 X kg)(2.2 X 10 6 m/s) = 3.3 X m or 0.33 nm

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DeBroglie Wavelength: Ex 3 Calculate the wavelength of an electron that has been accelerated through a potential difference of 100 V V = PE(PE =KE) q V = 1 mv 2 2 q v 2 = 2qV/m

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v = (2qV/m) 1/2 v=[(2)(1.602 X C)(100V)/(9.11 X kg)] 1/2 v = 5.9 X 10 6 m/s = h mv = (6.626 X J s) (9.11 X kg)(5.9 X 10 6 m/s) = 3.3 X m or 0.33 nm

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Electron Microscope Electron’s wavelength is smaller than light Magnetic focusing

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DNA

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Line Spectra Discharge tube –Low density gas (acts like isolated atoms) –Run a high voltage through it –Light is emitted Light is emitted only at certain (discrete) wavelengths Gases absorb light at the same frequency that they emit

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Hydrogen Helium Solar absorption spectrum

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Explaining the Lines

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Lyman Series 1 =R n 2 Balmer Series 1 =R n 2 Paschen Series 1 =R n 2

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Bohr Model: Hydrogen

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Electrons orbit in ground state (without radiating energy) Classically, electrons should radiate energy since that are accelerating because they are changing directions Jumps to excited state by absorbing a photon Returns to ground state by emitted a photon

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Bohr Model

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hf = E e - E g

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Bohr’s Equation Worls only for H and other 1- electron atoms (He +, Li 2+, Be 3+, etc…) Energy of ionized atom is set at 0 Orbital energies are below zero

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E n = eV n 2 E n = Energy of an orbital eV= Orbital of hydrogen closest to nucleus n = Number of the orbital (1 eV = 1.60 X J)

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Bohr’s Equation: Ex 1 Calculate the energy of the first three orbitals of hydrogen E n = eV n 2 E 1 = eV 1 2 E 1 = eV

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E 2 = eV 2 2 E 2 = -3.4 eV E 3 = eV 3 2 E 3 = eV

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Bohr’s Equation: Ex 2 What wavelength of light is emitted if a hydrogen electron drops from the n=2 to the n=1 orbit? E 1 = eV E 2 = -3.4 eV E = (-3.4eV eV) = 10.2 eV E = (10.2 eV)(1.60 X J/eV) = 1.63 X J E = hf f = c/

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E = hc/ = hc E = (6.626 X J s)(3.0 X 10 8 m/s) (1.63 X J) = 1.22 X m = 122 nm (UV)

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Bohr’s Equation: Ex 3 Calculate the wavelength of light emitted if a hydrogen electron drops from the n=6 to the n=2 orbit? (ANS: 410 nm (violet))

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Bohr’s Equation: Ex 4 Calculate the wavelength of light that must be absorbed to exite a hydrogen electron from the n=1 to the n=3 orbit? (ANS: 103 nm (UV))

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Bohr Model: Other Atoms E n = (Z 2 )(-13.6 eV) n 2 Z= Atomic number of the element (H=1, He=2, etc)

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Other Atoms: Ex 1 Calculate the ionization energy of He +. This is the energy needed to move an electron from n=1 to zero. E 1 = (2 2 )(-13.6 eV) 1 2 E 1 = eV

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Other Atoms: Ex 2 What wavelength of light would be required to ionize He + E 1 = eV 8.70 X J E = hf E = hc/ = hc = (6.626 X J s)(3.0 X 10 8 m/s) E(8.70 X J) = 22.8 nm

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Other Atoms: Ex 3 Calculate E 1 for a Li 2+ ion. E 1 = (3 2 )(-13.6 eV) 1 2 E 1 = eV

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Other Atoms: Ex 4 What wavelength of light would be emitted from a n=3 to n=1 transition in He + ? (ANS: = 34.2 nm)

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Wave/Particle Duality DeBroglie Each e - is actually a standing wave Only certain wavelengths produce resonance

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Forbidden Zone

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Circumference = 2 r 2 r n = n = h mv mvr n = nh 2

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