Making Light
• How do we make light?
Making Light
• How do we make light?
– Heat and Light: Incandescent Lighting
(10-20% efficient)
– Atoms and Light: Fluorescent Lighting
(40-60% efficient)
– Semiconductors and Light: Light Emitting
Diodes (LEDs) (60%-80% efficient)
We’ll consider Heat and Light first. Later in this
part we will consider Atoms and Light.
Blackbody Radiation:
• What is a blackbody?
A BLACK object absorbs all the light incident
on it.
A WHITE object reflects all the light incident
on it, usually in a diffuse way rather than in
a specular (mirror-like) way.
Blackbody Radiation:
• The light from a blackbody then 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.
Blackbody Radiation:
• If very hot objects glow (such as the filaments of
light bulbs and electric burners), do all warm
objects glow?
• Do we glow? (Are we warm? Are you HOT?)
Blackbody Radiation:
• What are the parameters associated with
the making of light from warm objects?
Blackbody Radiation:
• What are the parameters associated with
the making of light from warm objects?
– Temperature of the object.
– Surface area of the object.
– Color of the object ? (If black objects absorb
better than white objects, will black objects
emit better than white objects?)
Blackbody Radiation:
• Consider the following way of making your
stove hot and your freezer cold:
Blackbody Radiation:
Put a white object in an insulated and
evacuated box with a black object. The
black object will absorb the radiation from
the white object and become hot, while the
white object will reflect the radiation from
the black object and become cool.
Put the white object in the freezer, and the
black object in the stove.
Blackbody Radiation:
• Does this violate Conservation of Energy?
Blackbody Radiation:
• Does this violate Conservation of Energy?
NO
• Does this violate the Second Law of
Thermodynamics (entropy tends to
increase) ?
Blackbody Radiation:
• Does this violate Conservation of Energy? NO
• Does this violate the Second Law of
Thermodynamics (entropy tends to
increase) ? YES
• This means that a good absorber is also a
good emitter, and a poor absorber is a poor
emitter. 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 of the object, 
Blackbody Radiation:
• Is the  for us close to 0 or 1?
(i.e., are we white or black?)
We emit light in the IR, not the visible.
So what is our  for the IR?
Blackbody Radiation:
So what is our  for the IR?
Have you ever been near a fire on a cold
night?
Have you noticed that your front can get hot
at the same time your back can get cold?
Can your hand block this heat from the fire?
Is this due to convection or radiation?
Blackbody radiation:
• For humans in the IR, we are all fairly good
absorbers (black). An estimated value for 
for us then is about .97 .
Blackbody Radiation:
Experimental Results
• At 310 Kelvin, only get IR
Intensity
per
wavelength
(log scale)
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
per
wavelength
(log scale)
UV
blue
yellow
wavelength
red
IR
Blackbody Radiation:
Experimental Results
Itotal = Ptotal/A = T4 , or Ptotal = AT4
where  = 5.67 x 10-8 W/m2 *K4
peak = b/T where b = 2.9 x 10-3 m*K
Intensity
per
wavelength
(log scale)
UV
blue
yellow
wavelength
red
IR
Blackbody Radiation:
Example
• Given that you eat 2000 Calories/day,
your power output is around 100 Watts.
• Given that your body surface temperature is
about 90o F , and
• Given that your surface area is about
1.5 m2,
Blackbody Radiation:
Example
• Given Ptotal = 100 Watts
• Given that Tbody = 90o F
• Given that A = 1.5 m2
WHAT IS THE POWER EMITTED VIA
RADIATION?
Blackbody Radiation:
Example
• Pemitted = AT4
–
–
–
–
 = .97
= 5.67 x 10-8 W/m2 *K4
T = 273 + (90-32)*5/9 (in K) = 305 K
A = 1.5 m2
Pemitted = 714 Watts
(compared to 100 Watts generated!)
Blackbody Radiation:
Example
• need to consider power absorbed at room T
• Pabsorbed = AT4
–
–
–
–
 = .97
= 5.67 x 10-8 W/m2 *K4
T = 273 + (90-72)*5/9 (in K) = 295 K
A = 1.5 m2
Pabsorbed = 625 Watts
(compared to 714 Watts emitted!)
Blackbody Radiation:
Example
Total power lost by radiation =
714 W - 625 W = 89 Watts
(Power generated is 100 Watts.)
Power also lost by convection (with air)
and by evaporation.
Blackbody Radiation:
Example
• At colder temperatures, our emitted power
stays about the same while our absorbed
power gets much lower. This means that
we will get cold unless
– we generate more power, or
– our skin gets colder, or
– we reflect the IR back into our bodies.
• Use metal foil for insulation!
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.
• From thermodynamics, we have the
equipartition of energy: Each mode on
average has an energy proportional to the
Temperature of the object.
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
The standing wave theory and the
equipartition of energy theory together
predict that the intensity of light should
increase with decreasing wavelength:
This work 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
experiment
per
wavelength
wavelength
Blackbody Radiation:
Planck’s idea
• Need to turn the 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 (see page 5 of Study Guide
for Part 3), we have the Boltzmann probability
distribution function: P(E) = Ae-E/kT .
If we assume that energy is continuous, then the
summation can become an integral:
BOLTZMANN DISTRIBUTION
Probability of one atom having n units of
energy is based on equal likelihood of any
possible state. Following is a listing of all
possible states for two cases.
BOLTZMANN DISTRIBUTION
CASE I: four atoms having three units of energy:
ABCD ABCD ABCD ABCD ABCD ABCD
3000
0300
0030
0003
2100 1200 1020 1002 1110
2010 0210 0120 0102 1101
2001 0201 0021 0012 1011
0111
Summary of results (any order):
(3000) 4
(2100) 12
(1110) 4
Total number of different cases: 4+12+4=20
BOLTZMANN DISTRIBUTION
Case I: Prob of atom A having n of 3 units:
P(3) = 1/20 = .05
P(2) = 3/20 = .15
P(1) = 6/20 = .30
P(0) =10/20 = .50
Note that this probability distribution is
different than saying atom A has a (¼)
chance of getting each of the three units –
this would be (¼)3 = .016 . Why?
Boltzmann Distribution vs
Binomial Distribution
The probability distribution, called the
Binomial Distribution, that says each of 4
atom has a ¼ chance of getting each unit of
energy allows the chance that the four atoms
will accumulate more or less than the total
amount of 3 units of energy available.
BOLTZMANN DISTRIBUTION
CASE II: four atoms having five units of energy:
Prob of atom A having:
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
Plot of P(E) vs E
P(E) vs E
0.4
3 units
Series1
0.2
5 units
Series2
E
6
5
4
3
2
1
0
0
P(E)
0.6
Plot of E*P(E) vs E
P(E) and E*P(E)
Series1
P(E)
E*P(E)
Series2
0.6
0.4
E
3
2.5
2
1.5
1
0.5
0.2
0
0
E*P(E)
1
0.8
Blackbody Radiation:
Planck’s idea
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
E*P(E)
is more than area under green.
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.
The mathematical details of the actual summation
are considered in PHYS 447 (Modern Physics).
Blackbody Radiation:
Planck’s idea
To get the curve to fall at small wavelengths
(big frequencies) Planck tried the simplest
relation:
E = (constant) * f
since we need to decrease the average energy
per mode more as the wavelengths get
smaller - and the frequency gets bigger.
Blackbody Radiation:
Planck’s idea
Planck found that he could match the curve
and DERIVE both 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 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 energy 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: f = 1 MHz gives hf = 6.63 x 10-34 J*s
* 1 x 106 /s = 6.63 x 10-28 J = 4 x 10-9 eV
• red photon: = 700 nm, f = c/3x108 m/s / 7x10-7 m =
4.3 x 1014 Hz,
red photon energy = 1.78 eV
• blue: = 400 nm; photon energy = 3.11 eV .
• x-ray: = 1 nm; photon energy = 1,244 eV .
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.