work function Φ
Inside
metal
Now you change the
frequency to violet light
without changing the # of
photons per second.
Electrons over large range of energy have equal
chance of absorbing photons.
photoelectric_en.jar
Potential energy
of electrons
Electrons over large range of energy have about equal
chance of absorbing photons.
Say you shine blue light on
Ephot
a metal plate the metal
emits n electrons per sec.
Ephot
Electron potential
energy
Q1:
Ephot
work function Φ
metal
What happens to the
number of electrons/sec.
coming out of the metal?
a. fewer electrons / sec.
b. same # of electrons/sec but electrons are faster
c. more electrons/sec.
d. not enough information
e. no change because light is a wave
absorb blue light and have enough energy to leave
absorb blue light, but don’t come out
so the more energy the light has, the more electrons that come
out, until so much energy that every electron comes out.
(violet and ultraviolet would not be very different in this case)
Typical energies
Photon Energies:
Each photon has: Energy = Planks constant * Frequency
(Energy in Joules)
Application of the PE effect:
The photo multiplier tube (PMT)
(Energy in eV)
E=hf=(6.626*10-34 J-s)*(f in Hz)
E=hc/λ = (1.99*10-25 J-m)/(λ m)
Red Photon: 650 nm
c. more electrons come out with violet
E=hf= (4.14*10-15 eV-s)*(f in Hz)
E= hc/λ = (1240 eV-nm)/(λ nm)
Ephoton = 1240 eV-nm = 1.91 eV
650 nm
Electron
Work functions of metals (in eV):
Aluminum
4.08 eV
Cesium
2.1
Lead
4.14
Potassium
2.3
Beryllium
5.0 eV
Cobalt
5.0
Magnesium
3.68
Platinum
6.35
Cadmium
4.07 eV
Copper
4.7
Mercury
4.5
Selenium
5.11
Calcium
Carbon
2.9
4.81
Gold
Iron
5.1
4.5
Nickel
Niobium
5.01
4.3
Silver
Sodium
4.73
2.28
Uranium
3.6
Zinc
4.3
Q2:
Photomultiplier tubes: An application of photoelectric effect.
Most sensitive way to detect visible light, see single photons
Q3:
V
Metal plate
big voltage
current
electron multiplier:
gives pulse of
current for each
photoelectron
1
2
3
4
5
Time (millisec)
What would be the best
choice of these materials to
detect visible light?
a. Platinum
Φ = 6.35 eV
b. Magnesium
= 3.68 eV
c. Nickel
= 5.01 eV
d. lead
= 4.14 eV
e. Sodium
= 2.28 eV
A photon at 300 nm kicks out an electron with a maximum
initial kinetic energy KEλ=300nm. If the wavelength is halved
to 150 nm, and the photon hits an electron in the same
metal, what would be the maximum possible initial kinetic
energy KEλ=150nm of the released electron?
a. less than ½KEλ=300nm.
b. ½KEλ=300nm
c. = KEλ=300nm
d. 2 x KEλ=300nm
e. more than 2 x KE λ=300nm
(remember hill/kicker analogy, draw pictures to reason out
answer, don’t just pick answer without careful reasoning)
Q4:
KE = photon energy-energy to get out
= hf – Φ
KE300
If λ is ½ then, f twice as big, Ephot150 = 2 hf300
V
Old KE300 = hf300 - energy to get out
Shine in light of 300 nm. The most energetic electrons come
out with initial kinetic energy KEλ=300nm. A voltage difference of
1.8 V is required to stop these electrons. What is the work
function Φ for this metal? (e.g. the minimum amount of energy
needed to kick electron out of metal?)
New KEnew= 2hf300 - energy to get out
Energy
so KEnew is more than twice as big.
hf150
Remember: The energy of a photon
is hf (= hc/λ).
In units of eV you can write:
E= hc/λ = (1240 eV-nm)/(λnm)
a. 1.2 eV
b. 1.8 eV
c. 4.1 eV
d. 5.9 eV
e. none of the above
hf300
KE300
(Your answer should be within ~0.1eV)
KE300
This is the end of our discussion
about the PE effect
V
CQ: Shine in light of 300 nm, most energetic electrons come out with kinetic energy,
KE300. A voltage diff of 1.8 V is required to stop these electrons. What is the work
function Φ for this plate? (e.g. the minimum amount of energy needed to kick e out of
metal?)
Energy is conserved so:
a. 1.2 eV
b. 2.9 eV Ephot= energy need to exit (Φ) + electron’s left over energy
c. 6.4 eV
d. 11.3 eV so
Φ=E
– electron’s energy
phot
e. none
When electron stops, all of initial KE has been
of the
converted to electrostatic potential energy:
above
electron energy = q*∆V = e x 1.8V = 1.8 eV, and
Ephot = 1240 eV nm/300 nm = 4.1 eV.
Next:
More about photons and
the particle-wave duality
All these fifty years of conscious brooding have
brought me no nearer to the answer to the
question, 'What are light quanta?' Nowadays
every Tom, Dick and Harry thinks he knows it, but
he is mistaken! (Albert Einstein, 1954)
So Φ = 4.1eV - 1.8 eV = 2.3 eV
Electromagnetic Waves and Photons
are describing the same thing!
Common confusion:
Recap: Intensity ∝ Emax2 but does not depend on frequency!
Need to have a model where these views/perspectives all fit together
and are consistent with each other!
X
vs.
X
Interference was definitive test that light is a wave!
“Why do higher frequency gamma rays carry more energy (hf)
than lower frequency radio waves, but frequency has nothing to
do with intensity?”
Do not confuse the energy carried by a beam of light
with the energy of a single quantum particle of light!
1. Light, Radio waves, X-rays are all electromagnetic waves!
2. The electromagnetic wave is made up from lots of photons.
3. Photons could be thought of as mini E/M wave segments
(each has a specific energy hf and a wavelength c/f )
Electromagnetic
Wave
E
Travels
Straight
Made up from
photons
lots of photons
This picture can help explain wave view and particle view.
“Particle” = little chunk of the electromagnetic wave.
Energy of photon (hf) is in its oscillating E and B fields.
(Sometimes it also helps to think of a photon as a tiny particle/energy packet).
When is it important to think
about particle aspect of light?
Only if your detector is good enough to see individual
photons!
If you think of photons as particles you probably
automatically think of them as perfectly localized - like a
tiny billiard ball at a coordinate (x, y, z).
Watch out! This is not a good analogy for photons!!
• Billiard balls never produce a diffraction pattern
• Billiard balls have no measurable wavelength/frequency
• Billiard balls cannot go through two slits at the same time
(photons can; electrons too! Will show later)
Answers to clicker questions
Q1: C
Q2: E
Q3: E
Q4: E
Examples where important to think about particle behavior:
Photoelectric effect: Individual electrons popping out of metal
Lasers: Electrons in atoms transitioning between energy levels
Molecular bonds: Chemical bonds breaking with light
Quantum optics: Shot noise in a beam of light.
Examples where we don’t need to think about particle behavior
(detection system isn’t good enough and wave behavior is easier):
Heating: Energy absorbed in microwave or by black asphalt.
Optics: Light bending through lenses or passing through slits
Laser beam: Treat it just like a beam of lightT (Understanding
the inner workings of a laser requires photon picture.)