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Chapter 38: Photons, electrons and atoms

Light behaves like a wave:

• Interference• Diffraction• Refraction• Polarization

Interference

Diffraction

Refraction

Polarization

38-1: Emission and absorption of light

Emission line spectrumContinuous spectrum

Classical optics cannot explain these observations!

c = f

38-2: Photoelectric effect

Einstein won the Nobel Prize for this work in 1922!

Work function of a metal () :The minimum amount of energy an electron must gain so that it can be ejected from the metal surface.

No e-’s are ejected if f < threshold frequency.

1 eV = energy given to an e- by accelerating it through 1 V

Measure the e-’s maximum KE by varying the voltage, find what “stopping potential” V0 makes the e-’s stop (KE = 0).

Wtot = -eV0 = K = Kfinal - Kinitial

= 0 - Kmax

Kmax = eV0

Planck’s constant h = 6.626 x 10-34 J-s= 4.136 x 10-15 eV-s

When UV light of = 254 nm falls on a clean copper surface, the potential difference required to stop the photoelectrons is 0.181 V.

a) What is the work function for the surface? Compare to the known value.

b) What is the photoelectric threshold for this surface?

Photoelectric effect question

Suppose a metal has a work function of 4.5 eV. You shine light with an energy of 5.2 eV on the metal. What is the potential difference required to stop the current from flowing?

1) 0.7 V2) 4.5 V3) 5.2 V4) 9.7 V5) None of the above

Photoelectric effect question

You shine light with an energy of 4.3 eV on an unknown metal. You find that a potential difference of 1.2 V is required to reduce the current to zero. How much kinetic energy do the photoelectrons have?

1) 4.3 V2) 1.2 V3) 5.5 V4) 3.1 V5) You need to know the work function to solve the

problem

38.3 Atomic Line Spectra and Energy Levels

What causes the different colors?

Each elements has its own unique spectrum

Atoms have internal energy

The Balmer series for hydrogen: Visible light

electrons falling to n=2

Balmer’s formula (nf = 2) :1 / = R (1/nf

2 - 1/ni2)

R = Rydberg constant = 1.097 x 107 m-1

Hydrogen emission spectrum

The hydrogen atom

Energies are < 0 because we set E = 0 when atom is ionized. All other energies must be < 0 (like potential energy)

You can use a simple formula like Balmer’s only for atoms with one electron

• Hydrogen• Positronium• Singly ionized helium (He+)• Doubly ionized lithium (Li2+)

For other atoms, things get weird!

Energy levels and transitions of the many-electron atom: Sodium

Quantum states of the valence electron

An atom will absorb a photon only if the photon has exactly the right energy to “excite” a transition

Absorption spectrum

Very high resolution absorption spectrum tells us what the sun’s atmosphere is made of

38.4 The Nuclear Atom

Rutherford’s experiment

The nucleus is not just small, it is REALLY small!

Radius = 7x10-15 m Radius = 7x10-14 m

38.5 The Bohr Model

•Classical physics predicts that the electron should spiral into the nucleus•Cannot explain emission spectra

The Bohr model:•The e- stays in certain stable orbits, emits no radiation unless it jumps to a lower level

•The angular momentum of the e- is quantized

•the attaction between p and e- provides the centripetal acceleration

n = principal quantum number

From Coulomb’s law, the force between the proton and electron is

F = 1

40

q1 q2 r2

This is the centripetal force, mv2 / r

Where q1 = q2 = e for the hydrogen atom

Bohr radius a0 = 0h2 / me2 = 5.29 x 10-11 m

So when the electron is in any energy level n:

KE of the electron in the nth level: Kn = 1/2 mv2 = _____

-1

40

e2 r

PE of the electron in the nth level: Un =

Total energy En = Kn + Un = ???

Compare this with

Reduced mass: the nucleus is not infinite in mass, Bohr model is off by 0.1%

mr = m1 + m2

m1 m2

Positronium

The Bohr approximation works well for most atoms, because the nucleus is so massive compared to the electron. Not so for positronium.

What is the reduced mass in the positronium “atom”?

1. The mass of the electron2. 2x the mass of the electron3. 4x the mass of the electron4. 1/2 the mass of the electron5. 1/4 the mass of the electron6. No clue!

What does this do to the energy levels (and resulting frequencies of light) of positronium?

1. They are half of that for hydrogen 2. They are 2x that for hydrogen 3. They are 1/4 that for hydrogen 4. They are 4x that for hydrogen

Ionized Helium is also a 1-electron atom

Why is the emission spectrum of ionized helium similar to that of hydrogen?

1. Because hydrogen and helium are similar chemically

2. Because several of the energy levels of hydrogen and helium are the same

3. Because hydrogen and helium have similar atomic masses

4. It is a total coincidence

38.6 The Laser

Light Amplification through Stimulated Emission of Radiation

absorption emission

A photon can be emitted•Spontaneously•Stimulated by a collision•Stimulated by another photon

An atom can be excited •Due to a collision•Due to absorbing a photon

Nobel prize given to Charlie Townes!

Population inversion: more atoms in high energy state than low, without the high temperatures. Non equilibrium state!

nex

ngr =

exp(-Eex/kT)

exp(-Egr/kT)

Ratio of states: Maxwell-Boltzmann distribution

= exp(-(Eex -Egr)/kT

How a laser works

The electromagnetic spectrum

Maser: Naturally occurring!

38-7. X-ray production and scattering

=103-106 V

Roentgen discovered X rays in 1895.

•Not affected by direction of E field: No charge

•No mass either

•Must be a photon!

=103-106 V

As the e- are “braked” to rest, two things happen:1. their KE becomes photons of all energies: Bremsstrahlung

(braking radiation). Does not depend on target material!2. Some e- of the target material become excited and

produce x rays as they decay to lower states

Classical physics cannot explain this!

=

There should be an equal sign here!

Relationship between voltage and max photon energy:

The process by which x-rays are generated in this way can be compared to the photoelectric effect. How?

1. It is basically the same as the photoelectric effect2. it is the opposite of the photoelectric effect

Compton scattering: direct confirmation of the quantum nature of x rays

Photons lose energy as a result of scattering off of an electron in the target material. Here m = electron mass.

The missing energy goes into the recoil of a loosely bound electron

This peak is due to Compton scattering

38.8 Continuous spectra: Blackbody radiation

Spectral emittance

Intensity distribution I() = # of photons at some wavelength

The temperature of a blackbody uniquely determines the wavelength at which most of the photons are emitted:

= total intensity emitted over all wavelengths

I = total number of photons at all energies

= the area under the curve = the integral of I() over all

wavelengths 0--> infinity

38-9 Wave-particle duality

Particles:

:waves