Dual Nature of Light
Is light a wave or a particle?
Wave Properties
• -Diffraction
• -Interference
• -Polarization
Diffraction
Constructive &
Destructive Interference
Polarization
Energy
Wave E increases with A2/Intensity.
Studies of Wave E
• Planck – color (f, l) vs. T.
• As T inc. , f inc, (l decr)
Radiation & Temperature
Hot Objects Emit Waves
Intensity/Brightness
Problem:
Classical physics could not accurately
predict f vs. Temperature
Max Planck related f to T.
Light (EM) E, is quantized--it can only take on certain
whole number values.
E comes in little "chunks" of f x a constant now called
Planck's constant, h:
•
•
•
•
EM radiation waves
chunks
quanta
photons
Can calculate E in EM waves units
quanta or photons based on frequency.
E = hf.
h is Plank’s constant 6.63 x 10-34 Js.
E is energy in Joules
f is frequency of radiation
Show that if E = hf,
E = hc.
l
•
•
•
•
•
•
For waves, v = fl.
Rearrange f = v/l.
Vacuum/air EM v = c (3 x 108m/s).
f = c/l.
E = hf f = c/l.
E = hc.
l
Ex 1. Each photon of a certain
color light has an energy of 2.5
eV. What is the frequency of
and color of the light?
Solution:
E = hf
f = E/h
convert eV to Joules.
(2.5 eV)(1.6 x 10-19J/eV) = 6.03 x 1014 Hz
6.626 x 10-34 J s
Green Light
2. The energy of a certain photon is 2.9 eV.
What type of wave is it? Be specific.
• 2.9 eV x 1.6 x 10-19 J = 4.64 x 10-19 J.
eV
E=
hf
(4.64 x 10-19 J) = (6.63 x 10-34 Js) f
f = 7 x 10 14Hz
Violet Light
Finish Ex Sheet
• Hwk : Text Read 830 – 833 Do pg 833 #1-4
and 839 #2, & 856 # 2-4, 9.
Do Now: A photon of light has energy =
2.072 eV. What color is it?
•
•
•
•
•
•
2.072 eV (1.6 x 10-19 J/eV).
3.3152 x 10-19 J
E = hf.
(3.3152 x 10-19 J) = (6.63 x 10-34 Js)f
f = 5.00 x 1014 Hz.
Orange
So EM Radiation comes in chunks,
maybe it’s not waves.
• All objects above 0 K radiate EM waves as E.
• Hotter = more E = higher freq. (different color)
• E = hf (J).
What is the evidence?
• Photoelectric effect.
http://phet.colorado.edu/en/simulation/photoelectric
Photoelectric Effect
When EM waves shine on a metal surface, the E
in wave may be enough to kick out surface e-.
Materials that emit e- in this way are called
photoemissive. The ejected e- are called
photoelectrons.
Phet observations
•
•
•
•
•
Use higher Amplitude/Intensity/brightness –
more e- fly off w same speed.
Current increases (A, C/s)
Increased f
e- fly off faster w higher KE.
http://phet.colorado.edu/en/simulation/photoelectric
Classical Mechanics cannot explain why
increasing A or exposure time does not
increase photoemission. After all:
• Boat would be tossed higher & faster
with increased wave amplitude.
• But not ejected e-.
Einstein: Maybe EM waves quantized too –
photons.
• Increasing f, increases E of each photon.
• Increasing Intensity (A) increases number of
photons hitting more e- so more fly out.
• Envision EM as little chucks. High f are heavier.
• http://phet.colorado.edu/en/simulation/photoelectric
Photoemission only works with metals
with weakly bound e-.
Each metal, has a threshold frequency fo =
lowest f that will free an e-.
Light frequencies below the fo eject no e-, no
matter how intense or bright the light.
Light frequencies above the fo eject e-, no
matter how low the A (how dim).
A metal has a threshold frequency fo in the
blue light range.
1. What will happen if very bright red light
is shone upon the metal?
a) No e- will be emitted
b) more e- will be emitted
c) The emitted e- will have less energy.
Einstein confirmed EM waves come in
quanta or photons, each has E =hf.
Higher-frequency photons have more
E, so they can knock out e- & make the
e- fly out faster with greater KE.
2. A metal has a threshold frequency in
the blue light range.
What will happen if UV light is shone
upon the metal?
a) nothing
b) the emitted e- will have more
energy (KE)
c) more e- will be emitted with the
same energy.
Increasing the intensity/ampl, increases the
rate of e- emission or the current; the
number of e- ejected, but each e- won’t gain
any extra energy.
Only increased f increases E of
photoelectrons.
A metal has a threshold frequency in
the blue light range.
3. What will happen if the blue light is
made twice as bright?
a) nothing
b) the emitted e- will have more energy
(KE)
c) more e- will be emitted with the same
energy.
All EM waves can be described as quanta or
photons. The energy carried by photons is:
Ephoton = hf
or
Ephoton = hc/l.
(for photon traveling at speed of light).
This energy is absorbed by photo-emissive
materials.
The min. frequency to free e- is fo.
The min energy needed to free an eis called work function Wo, or F.
Metals have low Wo.
4. A certain metal has a work function
(Wo) of 1.7 eV. If photons of energy 3.0
eV are absorbed by the metal:
• a) No e- will be emitted at that energy.
• b) More e- will be emitted than would be
at the Wo.
• c) Higher KE e- will be emitted than
would be at Wo.
Summary:
•
•
•
•
•
•
•
•
•
•
EM waves as chunks of energy/photons travel at c.
Calculate the Energy J
E = hf, or
E = hc/l.
Evidence for photons – from Photoelectric Effect
Experiment –
f not A responsible for KE of ejected e-.
High f = high E, photon.
High A = high number of photons.
Photo-emissive materials have:
fo = min f to eject e- (Hz)
Wo= min E to eject e- (J)
Read Txt 830-835
Intro to Modern Physics Packet #1-14
Graph of Photoelectric
Experiment
• KE of photoelectron vs. frequency.
max KE of photo e- vs. f for metal. As f of EM
wave increases, KE increases, slope = h. F (work
function), is minimum energy needed to eject e-.
Work
function
State the work function & threshold
frequency of this metal
eV
0.4
0.0
0.4
0.8
1.0
2
3
4
5
6
x 1014 Hz
7
8
Wo & fo.
The threshold frequency fo, = min frequency
needed to eject e- so the min energy needed,
Wo in J is:
Wo = hfo.
Any photon E left over after the work
function, goes into KE of e-.
5. A particular metal has a threshold
frequency fo, of 5 x 1014 Hz.
What is its work function Wo in J & eV?
Wo = hfo.
3.3 x 10-19 J
2.07 eV
• Ephoton = hf is the total E available.
• Absorbed photon E splits between Wo & KE photo
e-, so total E of absorbed by e- is:
• Epho = Wo + KE.
• The maximum KE of ejected e- is:
• KEelc = Epho – Wo.
• Don’t forget Wo = hfo.
6: Photoelectric Effect:
Light having f = 1 x 1015 hz shines on a
sodium surface. The photoelectrons have
a maximum KE of 3 x 10-19 J.
Find the threshold frequency for sodium.
Photon Photoelectron.
Etot =
Wo + KE.
Etot – KE = Wo.
hf – KE = hfo.
fo = (hfphoton – KEmax)
(h)
change eV to Joules:
(1.86 eV) (1.6 x 10-19 J/eV) = 2.85 x 10-19 J
fo = (hfphoton – KEmax)/(h)
(6.63 x 10-34 Js)(1 x 1015 hz) - (2.85 x 10-19 J)
(6.63 x 10-34 Js)
fo = 5.5 x 1014 Hz.
Below this frequency no electrons will be
ejected.
In 1913-1914, R.A. Millikan did a series of
extremely careful experiments involving the
photoelectric effect. He found that all of his
results agreed exactly with Einstein's
predictions about photons, not with the
wave theory.
Einstein actually won the Nobel Prize for his
work on the photoelectric effect, not for his
more famous theory of relativity.
Some experimental results, like this one,
seem to prove that light consists of
particles; others insist, that it's waves.
We can only conclude that light is
somehow both a wave and a particle--or
that it's something else we can't quite
visualize, which appears to us as one or
the other depending on how we look at it.
Reg Hwk Intro Photoelectric Effect
Packet
• Hwk Text 834 – 837
• Do photo elec packet
Light Fantastic BBC part 3
58 min
• http://www.youtube.com/watch?v=VuGjo9
oNqao
Review of photoelec w german
accent 4.11
• http://www.youtube.com/watch?v=GpcWc5
KLVRo
• Photoelectric Effect Explained 6 min
• http://www.youtube.com/watch?v=0qKrOF
-gJZ4
Particle Properties of Waves
extend to conservation of
energy and momentum.
Photons may give up all or part of their
energy in collisions, but the sum of the
momentums and energy before must
equal the sum after.
Compton Effect
If light behaves like a particle, then a
collision btw photon & e- should be similar
to billiard balls colliding. Photons must
have momentum (p), & energy.
In collision of photons with particles (like
e-), conservation of energy & conservation
of momentum apply.
If the photon gives only part of its energy &
momentum to an e-, its momentum decreases
after the collision by the same amount as
absorbed by the electron.
Therefore, the frequency or energy of the
photon decreases. The wavelength increases.
pbefore = pafter.
E photon before = KEelc after. + E photon after
hfi = KEelc after + hff photon after
pphoton
= hf/c =
h/l. The
wavelength of the photon increases after
collision.
Matter has wave-like properties.
1924 Louis DeBroglie suggested that
since waves had particle properties,
matter might have wave properties.
It turns out that matter does have wave
properties which are inversely related to
the momentum of the particle.
For matter:
l =h/p
or
l = h/mv.
Since the mass of most objects is so large,
the wavelengths would be very small &
not measurable.
Electrons, however, do show diffraction &
other wave characteristics.