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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 Bohr De Broglie (Later Schrodinger/Heisenberg)

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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 10-3 mK = lpeakT

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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 10-9 m T = 2.90 X 10-3 mK lpeak T = 2.90 X 10-3 mK = 6000 K 500 X 10-9 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 10-3 mK = lpeakT l peak = 2.90 X 10-3 mK T l peak = 2.90 X 10-3 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 10-9m = 6 X10-7 m 1 nm c = lf f = c/l = (3X108 m/s)/(6 X10-7 m) = 5 X 1014 s-1 E = hf E = (6.626 X J s)(5 X 1014 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 10-9m = 5 X10-7 m 1 nm c = lf f = c/l = (3X108 m/s)/(5 X10-7 m) = 6 X 1014 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 1014 s-1) n = 2.5 X 1020

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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 Below a certain frequency, no electrons are emitted Greater intensity light produces more electrons Greater Frequency light produces no more electrons, but the come off with greater speed

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

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More intensity More photons More electrons ejected with same KE

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**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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Metal Work Function (eV) Na 2.46 Al 4.08 Cu 4.70 Zn 4.31 Ag 4.73 Pt 6.35 Pb 4.14 Fe 4.50

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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 (Wo) is 2.28 eV when illuminated with 410 nm light? 410 nm = 410 X 10-9 m or 4.10 X 10-7 m 2.28 eV X 1.60 X 10-19J = 3.65 X J 1 eV

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c = lf f = c/l f = c/(4.10 X 10-7 m) = 7.32 X 1014 s-1 hf = KE + W KE = hf – W KE = (6.626 X J s)(7.32 X 1014 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 (Wo) 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 (Wo) is 4.08 eV? 4.08 eV X 1.60 X 10-19J = 6.53 X J 1 eV

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hf = KE + W hf = 0 + W (looking for bare minimum) c = lf f = c/l hf = W hc = W l l = hc = (6.626 X J s)(3.0 X 108 m/s) W 6.53 X J l = 3.04 X 10-7 m = 304 nm (UV)

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**Photon/Matter Interactions**

Electron excitation (photon disappears) Ionization/photoelectric effect (photon disappears) Scattering by nucleus or electron 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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l’ = l + h (1 – cos q) moc l’ = wavelength of scattered photon l = wavelength of incident photon mo = rest mass of particle q = 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 0o? l’ = l + h (1 – cos q) moc l’ = 140 X10-9m + (6.626 X J s)(1 – cos 0) (9.11 X kg)(3 X 108m/s) l’ = 140 X10-9m + 0 l’ = 140 nm

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Compton Effect: Ex 2 What will be the wavelength of the X-rays scattered at 90o? l’ = l + h (1 – cos q) moc l’ = 140 X10-9m + (6.626 X J s)(1 – cos 90) (9.11 X kg)(3 X 108m/s) l’ = 140 X10-9m X m l’ = 142 nm

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Compton Effect: Ex 3 What will be the wavelength of the X-rays scattered at 180o? l’ = l + h (1 – cos q) moc l’ = 140 X10-9m + (6.626 X J s)(1 – cos 180) (9.11 X kg)(3 X 108m/s) l’ = 140 X10-9m X m l’ = 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 Wave Particle Prism Blackbody 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 E2 = p2c2 + m2c4 (consider a photon) E2 = p2c2 (photon has no mass) E = pc E = hf hf = pc c = lf

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**hf = plf p = mv (for a particle) hf = mvlf h = mvl l = 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 l = h mv l = (6.626 X J s) (0.20 kg)(15 m/s) l = 2.2 X m

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**DeBroglie Wavelength: Ex 2**

Calculate the wavelength of an electron moving at 2.2 X 106 m/s l = h mv l = (6.626 X J s) (9.11 X kg)(2.2 X 106 m/s) l = 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 mv2 2 q v2 = 2qV/m

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**v = (2qV/m)1/2 v=[(2)(1. 602 X 10-19 C)(100V)/(9**

v = (2qV/m)1/2 v=[(2)(1.602 X C)(100V)/(9.11 X kg)]1/2 v = 5.9 X 106 m/s l = h mv l = (6.626 X J s) (9.11 X kg)(5.9 X 106 m/s) l = 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 1 - 1 l 12 n2 Balmer Series l 22 n2 Paschen Series l 32 n2 **

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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 = Ee - Eg

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Bohr’s Equation Worls only for H and other 1-electron atoms (He+, Li2+, Be3+, etc…) Energy of ionized atom is set at 0 Orbital energies are below zero

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**En = -13. 6 eV n2 En = Energy of an orbital -13**

En = eV n2 En = 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 En = eV n2 E1 = eV 12

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**E2 = -13.6 eV 22 E2 = -3.4 eV E3 = -13.6 eV 32 E3 = -1.51 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? E1 = eV E2 = -3.4 eV DE = (-3.4eV eV) = 10.2 eV DE = (10.2 eV)(1.60 X 10-19J/eV) = 1.63 X 10-18J DE = hf f = c/l

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DE = hc/l = hc DE = (6.626 X J s)(3.0 X 108 m/s) (1.63 X 10-18J) l = 1.22 X 10-7 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**

En = (Z2)(-13.6 eV) n2 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. E1 = (22)(-13.6 eV) 12 E1 = eV

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Other Atoms: Ex 2 What wavelength of light would be required to ionize He+ E1 = eV X J E = hf E = hc/l = hc = (6.626 X J s)(3.0 X 108 m/s) E (8.70 X J) l = 22.8 nm

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Other Atoms: Ex 3 Calculate E1 for a Li2+ ion. E1 = (32)(-13.6 eV) 12 E1 = 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: l = 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 = 2pr 2prn = nl l = h mv mvrn = nh 2p **

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