- By
**jase** - Follow User

- 78 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about 'Early Quantum Mechanics' - jase

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

### Early Quantum Mechanics

Chapter 27

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

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

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

- 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

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

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

Photons: Ex 2

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

Photons: Ex 2a

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

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

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)

- 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)

- 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

- More photons
- More electrons ejected with same KE

- No more electrons ejected
- Electrons come off with greater speed (KE)

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

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

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

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

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

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

- Electron excitation (photon disappears)
- Ionization/photoelectric effect (photon disappears)
- Scattering by nucleus or electron
- Pair production (photon disappears)

- Photon is absorbed (disappears)
- Electron jumps to an excited state

- Ionization/Photoelectric Effect
- Photon is absorbed (disappears)
- Electron is propelled out of the atom

- 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

- 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

moc

l’ = wavelength of scattered photon

l = wavelength of incident photon

mo = rest mass of particle

q = angle of incidence

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

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

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

- 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

- 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)

- Cathode Ray Tube
- Charged particles produced (affected by magnetic field)

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)

- 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)

- Gold Foil Experiment
- Discovers nucleus (disproves Plum Pudding Model)
- Planetary Model

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

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

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

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

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

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’s wavelength is smaller than light
- Magnetic focusing

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

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

- 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

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

Calculate the energy of the first three orbitals of hydrogen

En = -13.6 eV

n2

E1 = -13.6 eV

12

E1 = -13.6 eV

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

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

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))

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

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 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

DeBroglie

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

Download Presentation

Connecting to Server..