Making Light
How do we make light?
Making Light
How do we make light?
– Heat and Light: Incandescent Lighting
(5% efficient at ~ 16 lumens/Watt)
– Atoms and Light:
• Fluorescent Lighting (20% efficient at ~ 70 lumens/Watt)
• LED’s (90% efficient at ~ 300 lumens/Watt)
We’ll review Heat and Light first.
Later in this part we will consider Atoms and Light.
Blackbody Radiation:
Heat loss by radiation was discussed in PHYS 201 in
the Thermo part.
The light from a blackbody is light that comes solely
from the object itself rather than being reflected
from some other source.
A good way of making a blackbody is to force
reflected light to make lots of reflections: inside a
bottle with a small opening. [Even if light loses only
5% on each reflection, after 25 reflections the light is down
to 28% of its original intensity.]
Blackbody Radiation:
Recall that a good absorber is also a good
emitter, and a poor absorber is a poor
emitter. We use the symbol  to indicate
the blackness (=1) or the whiteness (=0)
of an object.
Blackbody Radiation:
What are the parameters associated with the
making of light from warm objects?
– Temperature of the object, T.
– Surface area of the object, A.
– Color (black versus white) of the object, 
Blackbody Radiation:
Experimental Results
At 310 Kelvin, only get IR
Intensity
UV
blue
yellow
wavelength
red
IR
Blackbody Radiation:
Experimental Results
At much higher temperatures, get visible
- look at blue/red ratio to get temperature
Intensity
log scale
UV
blue
yellow
wavelength
red
IR
Experimental Results
The preceding graph is a little misleading in that the
scales for the Intensity axis are not the same for
the different temperatures.
For example, comparing peaks: the peak of the
T=300 K curve is about 160,000 times smaller
than the peak of the T=6,000 K curve.
At   9.7 microns (in the IR) which is the peak of
the 300 K curve, the 300 K curve is about 120
times lower than the 6,000 K curve at that same
wavelength.
Blackbody Radiation:
Experimental Results
Ptotal = AT4
where  = 5.67 x 10-8 W/m2 *K4
peak = b/T where b = 2.9 x 10-3 m*K
Intensity
log scale
UV
blue
yellow
wavelength
red
IR
Example
If you eat 2,000 calories per day, that is
equivalent to about 100 joules per second or
about 100 Watts - which must be emitted.
Let’s see how much radiation you emit when
the temperature is comfortable, say
75oF=24oC=297K, and pick a surface area,
say 1.5m2, that is at a temperature of
93oF=34oC=307K: Pemitted = AT4 =
(5.67x10-8W/m2K4)*(.97)*(1.5m2)*(307K)4 =
733 Watts emitted!
Example continued
But this is not the whole story: besides
emitting radiation, we receive radiation
from the outside: Pabsorbed = AT4 =
(5.67x10-8W/m2K4)*(.97)*(1.5m2)*(297K)4 =
642 Watts absorbed!
Hence, the net power emitted by the body via
radiation is: Pnet = 733 Watts - 642 Watts
= 91 Watts. The peak of this radiation is at:
peak = b/T = 2.9x10-3m*K / 307K = 9.5m
which is in the infrared (as expected).
Blackbody Radiation:
Wave Theory
Certain waves resonate in an object (due to
standing wave), such that n(/2) = L.
From this it follows that there are more small
wavelengths that fit than long wavelengths.
(see next slide for example)
Blackbody Radiation:
Wave Theory
n(/2) = L
Example: for L = 1 meter, we have the
following wavelengths that “fit”:
1 = 2 m; 2 = 1 m; 3 = .67 m; 4 = .50 m;
5 = .40 m; 6 = .33 m; 7 = .29 m; 8 = .25 m; etc.
For the range of ’s,
we have permitted
1 - 1.99 m;
1
.50 - .99 m (half the range size),
2
.25 - .49 m (half again the range size), 4
etc.
Blackbody Radiation:
Wave Theory
From thermodynamics, we have the
equipartition of energy: Each mode on
average has an energy proportional to the
Temperature of the object: Emode = kT,
where k = 1.38 x 10-23 Joules/Kelvin.
Blackbody Radiation:
Wave Theory
The standing wave theory and the
equipartition of energy theory together
predict that the intensity of light should
increase with decreasing wavelength:
This works very well at long wavelengths, but
fails at short wavelengths. This failure at
short wavelengths is called the ultraviolet
catastrophe.
Blackbody Radiation:
Wave Theory
wave theory: UV catastrophe
Intensity
log scale
experiment
wavelength
Blackbody Radiation:
Planck’s idea
• Need to turn the Intensity curve down when
 gets small (or frequency gets large).
• Keep standing wave idea and number of
modes.
• Look at equipartition theory and how the
energy per mode got to be kT (where k is
Boltzmann’s constant: k = 1.38 x 10-23 J/K).
Blackbody Radiation:
Planck’s idea
Eavg = Ei /1 = P(E)*E / P(E)
where P(E) is the probability of having energy, E.
From probability theory we have the Boltzmann
probability distribution function: P(E) = Ae-E/kT
(this is demonstrated in the next few slides).
If we assume that energy is continuous, then the
summation can become an integral, and we can get
a value for the average energy (which we do after
discussing the Boltzmann probability distribution).
BOLTZMANN DISTRIBUTION
Probability of one atom having n units of
energy P(En) = P(n*E1) is based on equal
likelihood of any atom having any of the n
units. Following is a listing of all possible
cases for 4 atoms (called A, B, C and D)
having three units of energy. After that we
will state the results of 4 atoms having five
units of energy.
BOLTZMANN DISTRIBUTION
CASE I:
four atoms having three units of energy:
ABCD ABCD ABCD ABCD ABCD ABCD
3000 2100 1200 1020 1002 1110
0300 2010 0210 0120 0102 1101
0030 2001 0201 0021 0012 1011
0003
0111
(3000) 4 (2100) 12
(1110) 4
BOLTZMANN DISTRIBUTION
Case I:
Probability of atom A having n units: P(n):
P(3) = 1/20 = .05
P(2) = 3/20 = .15
P(1) = 6/20 = .30
P(0) =10/20 = .50
Note that the sum of probabilities equals 1 as
it should:
P(n) = 20/20 = 1.00
BOLTZMANN DISTRIBUTION
Note that this distribution is different than the
distribution based on straight probability.
For example, it may seem that the probability of an
atom getting a unit of energy would be ¼ for each
of the three units available, so the probability that
one atom would get a unit of energy in all three
distributions would be (1/4)3 = .0156 (compared to
the Boltzmann probability of .0500). But we can’t have
all four atoms having 3 units of energy since that
would mean we would need 12 units of energy
when we only have 3 units total!
BOLTZMANN DISTRIBUTION
Likewise, the straight probability of one atom
getting no units of energy in each
distribution would be (3/4), so the straight
probability of one atom getting zero after
three distributions would be (3/4)3 = .422
(compared to the Boltzmann’s probability of
.500), but we can’t have all four atoms
having zero units since we do have 3 units
of energy that must be distributed.
BOLTZMANN DISTRIBUTION
CASE II: four atoms
Case I:
Probability of atom A having:
five units of energy:
three units of energy
P(5) = 1/56 = .018
P(4) = 3/56 = .054
P(3) = 6/56 = .107
P(2) =10/56 = .179
P(1) =15/56 = .268
P(0) =21/56 = .375
P(5) = .000
P(4) = .000
P(3) = .050
P(2) = .150
P(1) = .300
P(0) = .500
Plot of P(E) vs E
P(E) vs E
4 units
0.4
Series1
0.2
Series2
5 units
E
6
5
4
3
2
1
0
0
P(E)
0.6
P(E) vs E
• Note that both of these look approximately like
dying exponentials.
• Note that P(E) for large values of E gets bigger
and P(E) for small values of E gets smaller if
there is more energy to share.
• Note that since the sum of P(E) over all E gives 1,
the value of P(0) decreases as E gets bigger.
The function P(E) = Ae-E/kT is consistent with all of
these.
What is important in calculating the average energy
is the product of E and P(E). We look at this on
the next slide.
Plot of E*P(E) vs E
P(E) and E*P(E)
P(E)
Series1
0.6
0.4
Series2
0.2
0
E
3
2.5
2
1.5
1
0.5
E*P(E)
0
E*P(E)
1
0.8
Blackbody Radiation:
Sum goes to Integral
Eavg = LIME→0 [P(E) / P(E)] =


0
0
 E * P( E ) dE /  P( E ) dE
=


0
0
 E / kT
 E / kT
E
*
Ae
dE
/
Ae
dE


= Area under the curve / 1 = kT .
Blackbody Radiation:
Planck’s idea
Planck recalled that the SUM only became the
INTEGRAL if you let E go to zero.
Planck’s idea was NOT to let E go to zero.
If you require P(E) to be evaluated at the end
of each E, then the SUM will decrease as
E increases!
Blackbody Radiation:
Planck’s idea
As E gets bigger, Eavg gets smaller:
E*P(E) = A*E*e-E/kT . Area under red curve
is more than area under blue
is more than area under green.
E*P(E)
E
Blackbody Radiation:
Planck’s idea
It’s easy to see on the leading edge that as E
gets bigger, the total Energy under the
curve and hence the average energy gets
smaller. This is in fact confirmed by an
actual summation.
Eavg ↓ as E ↑
Blackbody Radiation:
Planck’s idea
To get the curve to fall at small wavelengths
Planck tried the simplest relation:
E α 1/ , or E = (constant) * f
since we need to decrease the average energy per
mode more as the wavelengths get smaller - and
the frequency gets bigger:
Eavg ↓ as E ↑ as  ↓ as f ↑ .
Blackbody Radiation:
Planck’s idea
Planck found that he could match the curve and
DERIVE the empirical relations:
– P = AT4 where  = 5.67 x 10-8 m2 *K4
– max = b/T where b = 2.9 x 10-3 m*K
with the simplest relation:
E = (constant) * f
if the constant = 6.63 x 10-34 J*sec = h.
The constant, h, is called Planck’s constant.
How to Make Light
The wave theory combined with the
equipartition of energy theory failed to
explain blackbody radiation.
Planck kept the wave idea of standing waves
but introduced E = hf, the idea of light
coming in discrete packets (or photons)
rather than continuously as the wave theory
predicted.
How to Make Light
From this theory we now have a way of
relating the photon idea to the wave idea for
color and type: E = hf .
– Note that high frequency (small wavelength) light
has high photon energy, and that low frequency
(large wavelength) light has low photon energy.
How to Make Light
E = hf
High frequency light tends to be more
dangerous than low frequency light
(UV versus IR, x-ray versus radio).
The photon theory gives a good account
of why the frequency of the light makes a
difference in the danger. Individual
photons cannot break bonds if their
energy is too low while big photons can!
Photons and Colors
• Electron volts are useful size units of energy
1 eV = 1.6 x 10-19 Coul * 1V = 1.6 x 10-19 J.
• radio photon: hf = 6.63 x 10-34 J*s * 1 x 106 /s
= 6.63 x 10-28 J = 4.15 x 10-9 eV
• red photon: f = c/3 x 108 m/s / 7 x 10-7 m =
4.3 x 1014 Hz,
red photon energy = 1.78 eV
• blue: = 400 nm; photon energy = 3.11 eV .
• X-ray photon: hf = 6.63 x 10-34 J*s * 1 x 1018 /s
= 6.63 x 10-16 J = 4,150 eV
Power and photons
Example
How many photons are emitted every second
from one watt of yellow light?
Power = Energy / time
= Energy per photon * number of photons / time
= hf * n/sec;
f=c, so f = c/
Power = (hc/) * n/sec
P = 1 Watt =
(6.63 x 10-34 J-s * 3 x 108 m/s / 5.5 x 10-7m) * n/sec;
n/sec = 2.8 x 1018 photons per second.
Making and Absorbing Light
• The photon theory with E = hf was useful
in explaining the blackbody radiation.
• Is it useful in explaining other experiments?
• We’ll consider next the photoelectric effect.
Photoelectric Effect
Light hits a metal plate, and electrons are
ejected. These electrons are collected in the
circuit and form a current.
light
ejected electron
A
- +
V
Photoelectric Effect
The following graphs illustrate what the wave
theory predicts will happen:
current
light
intensity
current
Voltage
current
frequency of light
Photoelectric Effect
We now show in blue what actually happens:
current
light
intensity
current
V-stop
Voltage
current
f-co
frequency of light
Photoelectric Effect
In addition, we see a connect between V-stop
and f above fcutoff:
V-stop
fcutoff
frequency
Photoelectric Effect
Einstein received the Nobel Prize for his
explanation of this. (He did NOT receive
the prize for his theory of relativity.)
Photoelectric Effect
Einstein suggested that light consisted of discrete
units of energy, E = hf. Electrons could either
get hit with and absorb a whole photon, or they
could not. There was no in-between (getting part
of a photon).
If the energy of the unit of light (photon) was not
large enough to let the electron escape from the
metal, no electrons would be ejected. (Hence, the
existence of f-cutoff.)
Photoelectric Effect
If the photon energy were large enough to
eject the electron from the metal (here, W is
the energy necessary to eject the electron),
then the following equation would apply:
hf = W + KE
The Energy of the photon absorbed (hf) goes
into ejecting the electron (W) plus any extra
energy left over which would show up as
kinetic energy (KE).
Photoelectric Effect
This extra kinetic energy (KE) would allow
the electron to climb up a “hill”, but the size
of the hill that the electron could climb up
would be limited to the extra kinetic energy
the electron had. By measuring the highest
hill, we could arrive at the extra energy of
the electron.
Hill sizes in electrical terms are in VOLTS:
KE = PE = qVstop.
Photoelectric Effect
Put into a nice equation:
• hf = W + e*Vstop
– where f is the frequency of the light
– W is the “WORK FUNCTION”, or the
amount of energy needed to get the electron out
of the metal
– Vstop is the stopping potential
• When Vstop = 0, f = fcutoff , and hfcutoff = W.
Photoelectric Effect - Example
Most metals have a work function on the
order of several electron volts. Copper has
a work function of 4.5 eV.
Therefore, the cut-off frequency for light
ejecting electrons from copper is:
hfcutoff = 4.5 eV, or
fcutoff = 4.5 x (1.6 x 10-19 C) x (1 V) / 6.63 x 10-34 J-sec
= 1.09 x 1015 Hz,
Photoelectric Effect - Example
or cutoff = c/ fcutoff , or
cutoff = (3 x 108 m/s) / (1.09 x 1015 cycles/sec)
= 276 nm (in the UV range)
Any frequency lower than the cut-off (or any
wavelength greater than the cut-off value) will
NOT eject electrons from the metal.
Photoelectric Effect
From Einstein’s equation:
hf = W + e*Vstop , we can see that the
straight line of the Vstop vs f graph should
have a slope of (h/e) . This gives a second
way of determining the value of h. [The
first was from fitting the blackbody curve.]
When we do this, we get the same value for
h that Planck did: 6.63 x 10-34 Joule*sec .
Computer Homework
The computer homework program on
Photons (Vol 5, #5) deals with both
Blackbody Radiation and the Photoelectric
Effect.
Compton Scattering
When light encounters charged particles, the
particles will interact with the light and
cause some of the light to be scattered.
light wave
electron
motion of electron
incident
photon
scattered
photon
electron motion of
electron
after hit
Compton Scattering
From the wave theory, we can understand that
charged particles would interact with the
light since the light is an electromagnetic
wave!
Compton Scattering
But the actual predictions of how the light
scatters from the charged particles does not
fit our simple wave model.
If we consider the photon idea of light, some
of the photons would “hit” the charged
particles and “bounce off”. The laws of
conservation of energy and momentum
should then predict the scattering.
Compton Scattering
From the theory of relativity, E = mc2, and
since photons have E = hf, they have mass
and so DO HAVE MOMENTUM as well as
energy. The scattered photons will have
less energy and less momentum after
collision with electrons, and so should have
a larger wavelength according to the
formula:
 = scattered - incident = (h/mc)[1-cos()]
Compton Scattering
 = scattered - incident = (h/mc)[1-cos()]
Note that Planck’s constant is in this relation
as well, and gives a further experimental
way of getting this value.
Again, the photon theory provides a nice
explanation of a phenomenon involving
light.
Compton Scattering
 = scattered - incident = (h/mc)[1-cos()]
Note that the maximum change in wavelength
is (for scattering from an electron)
2h/mc = 2(6.63 x 10-34 J-s) / (9.1 x 10-31 kg * 3 x 108 m/s)
= 4.86 x 10-12 m
which would be insignificant for visible light
( ≈ 10-7m) but NOT for x-ray and -ray light
( ≈ 10-10m and smaller).