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Early Quantum Mechanics Chapter 27. Three Major Discoveries Light is both a wave and a particle Electron Orbitals are Quantized The electron (and all.

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Presentation on theme: "Early Quantum Mechanics Chapter 27. Three Major Discoveries Light is both a wave and a particle Electron Orbitals are Quantized The electron (and all."— Presentation transcript:

1 Early Quantum Mechanics Chapter 27

2 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)

3 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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5 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

6 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

7 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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9 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

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

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

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

13 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

14 100 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

15 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)

16 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

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

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

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

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

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

22 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

23 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

24 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

25 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

26 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)

27 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)

28 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

29 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.

30 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

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

32 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

33 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

34 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)

35 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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37 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

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

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

40 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

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

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43 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

44 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

45 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

46 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

47 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

48 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

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

50 DNA

51 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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53 Hydrogen Helium Solar absorption spectrum

54 Explaining the Lines

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

56 Bohr Model: Hydrogen

57 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

58 Bohr Model

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60 hf = E e - E g

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62 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

63 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)

64 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

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

66 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/

67  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)

68 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))

69 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))

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

71 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

72 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

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

74 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)

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

76 Forbidden Zone

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


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