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Early Quantum Mechanics. Chapter 27. Three Major Discoveries. 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 = l peak T.

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wein s law
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

wein s law ex 1
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

wein s law ex 2
Wein’s Law: Ex 2

Suppose a star has a surface temperature of about 3500 K. Estimate the wavelength of light produced.

2.90 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

planck s quantum hypothesis
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 = 6.626 X 10-34 J s (Planck’s constant)
photons ex 1
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 10-34 J s)(5 X 1014 s-1) = 3.3 X 10-19 J

photons ex 1a
Photons: Ex 1a

Convert your answer from the previous problem to electron Volts (1 eV = 1.6 X 10-19 J)

photons ex 2
Photons: Ex 2

Calculate the energy of a photon of wavelength 450 nm (blue light).

photons ex 2a
Photons: Ex 2a

Convert your answer from the previous problem to electron Volts (1 eV = 1.6 X 10-19 J)

photons ex 3
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

slide14

100 Watts = 100 J/s (we are looking at 1 second)

E = nhf

n = E/hf

n = 100 J

(6.626 X 10-34 J s)(6 X 1014 s-1)

n = 2.5 X 1020

photoelectric effect einstein
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)
slide16

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
slide18

More intensity

    • More photons
    • More electrons ejected with same KE
slide19

Greater Frequency

    • No more electrons ejected
    • Electrons come off with greater speed (KE)
hf ke w
hf = KE + W

hf = energy of the photon

KE = Maximum KE of the emitted electron

W = Work function to eject electron

hf ke w ex 1
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 10-19 J

1 eV

slide23

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 10-34 J s)(7.32 X 1014 s-1) - 3.65 X 10-19 J

KE = 1.20 X 10-19 J or 0.75 eV

hf ke w ex 2
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

hf ke w ex 3
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 10-19 J

1 eV

slide26

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 10-34 J s)(3.0 X 108 m/s)

W 6.53 X 10-19 J

l = 3.04 X 10-7 m = 304 nm (UV)

photon matter interactions
Photon/Matter Interactions
  • Electron excitation (photon disappears)
  • Ionization/photoelectric effect (photon disappears)
  • Scattering by nucleus or electron
  • Pair production (photon disappears)
slide28

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
slide29

Pair Production

  • Photon closely approaches a nucleus
  • Photon disappears
  • An electron and positron are created.
  • Scattering
  • Photon collides with a nucleus or electron
  • Photon loses some energy
  • Speed does not change, but the wavelength increases
3 scattering compton effect
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
slide31

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

compton effect ex 1
Compton Effect: Ex 1

X-rays of wavelength 0.140 nm are scatterd from a block of carbon. What will be the wavelength of the X-rays scattered at 0o?

l’ = l + h (1 – cosq)

moc

l’ = 140 X10-9m + (6.626 X 10-34 J s)(1 – cos0)

(9.11 X 10-31 kg)(3 X 108m/s)

l’ = 140 X10-9m + 0

l’ = 140 nm

compton effect ex 2
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 10-34 J s)(1 – cos 90)

(9.11 X 10-31 kg)(3 X 108m/s)

l’ = 140 X10-9m + 2.4 X 10-12 m

l’ = 142 nm

compton effect ex 3
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 10-34 J s)(1 – cos 180)

(9.11 X 10-31 kg)(3 X 108m/s)

l’ = 140 X10-9m + 4.8 X 10-12 m

l’ = 145 nm (straight back)

4 pair production
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
principle of complimentarity
Principle of Complimentarity
  • Neils Bohr
  • Any experiment can only observe light’s wave or particle properties, not both
  • Different “faces” that light shows
the discovery of the electron thomson
The Discovery of the Electron (Thomson)
  • Cathode Ray Tube
  • Charged particles produced (affected by magnetic field)
slide39

Concluded that atom must have positive and negative parts

  • Electron – negative part of the atom
  • Only knew the e/m ratio
  • Plum Pudding Model
charge and mass of the electron millikan
Charge and Mass of the Electron (Millikan)
  • Oil drop experiment
  • Determines charge on electron (uses electric field to counteract gravity)
  • Quantized
  • e = 1.602 X 10-19 C
  • m = 9.11 X 10-31 kg
the nucleus rutherford
The Nucleus (Rutherford)
  • Gold Foil Experiment
  • Discovers nucleus (disproves Plum Pudding Model)
  • Planetary Model
wave nature of matter
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

slide44

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

debroglie wavelength ex 1
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 10-34 J s)

(0.20 kg)(15 m/s)

l = 2.2 X 10-34 m

debroglie wavelength ex 2
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 10-34 J s)

(9.11 X 10-31 kg)(2.2 X 106 m/s)

l = 3.3 X 10-10 m or 0.33 nm

debroglie wavelength ex 3
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

slide48

v = (2qV/m)1/2

v=[(2)(1.602 X 10-19 C)(100V)/(9.11 X 10-31 kg)]1/2

v = 5.9 X 106 m/s

l = h

mv

l = (6.626 X 10-34 J s)

(9.11 X 10-31 kg)(5.9 X 106 m/s)

l = 3.3 X 10-10 m or 0.33 nm

electron microscope
Electron Microscope
  • Electron’s wavelength is smaller than light
  • Magnetic focusing
line spectra
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
slide53

Hydrogen

Helium

Solar absorption spectrum

slide55

Lyman Series

1 = R 1 - 1

l 12 n2

Balmer Series

1 = R 1 - 1

l 22 n2

Paschen Series

1 = R 1 - 1

l 32 n2

slide57

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
bohr s equation
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
slide63

En = -13.6 eV

n2

En = Energy of an orbital

-13.6 eV = Orbital of hydrogen closest to nucleus

n = Number of the orbital

(1 eV = 1.60 X 10-19 J)

bohr s equation ex 1
Bohr’s Equation: Ex 1

Calculate the energy of the first three orbitals of hydrogen

En = -13.6 eV

n2

E1 = -13.6 eV

12

E1 = -13.6 eV

slide65

E2 = -13.6 eV

22

E2 = -3.4 eV

E3 = -13.6 eV

32

E3 = -1.51 eV

bohr s equation ex 2
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 = -13.6 eV

E2 = -3.4 eV

DE = (-3.4eV - -13.6 eV) = 10.2 eV

DE = (10.2 eV)(1.60 X 10-19J/eV) = 1.63 X 10-18J

DE = hf

f = c/l

slide67

DE = hc/l

  • = hc

DE

  • = (6.626 X 10-34 J s)(3.0 X 108 m/s)

(1.63 X 10-18J)

l = 1.22 X 10-7 m = 122 nm (UV)

bohr s equation ex 3
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))

bohr s equation ex 4
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))

bohr model other atoms
Bohr Model: Other Atoms

En = (Z2)(-13.6 eV)

n2

Z = Atomic number of the element (H=1, He=2, etc)

other atoms ex 1
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 = -54.4 eV

other atoms ex 2
Other Atoms: Ex 2

What wavelength of light would be required to ionize He+

E1 = -54.4 eV  8.70 X 10-18 J

E = hf

E = hc/l

  • = hc = (6.626 X 10-34 J s)(3.0 X 108 m/s)

E (8.70 X 10-18 J)

l = 22.8 nm

other atoms ex 3
Other Atoms: Ex 3

Calculate E1 for a Li2+ ion.

E1 = (32)(-13.6 eV)

12

E1 = -122.4 eV

other atoms ex 4
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)

wave particle duality
Wave/Particle Duality

DeBroglie

  • Each e- is actually a standing wave
  • Only certain wavelengths produce resonance
slide77

Circumference = 2pr

2prn = nl

l = h

mv

mvrn = nh

2p