Download Lecture 15

Electromagnetic waves carry energy
Emax
Electromagnetic waves carry energy
Emax=peak amplitude
X #1
c
Emax
X
X #2
E1max=E2max>E3max
f1 = f3 < f2
X #3
Which barrel will heat up the fastest?
a. 2>1>3
b. 1>2>3
c. 1=2>3
2
d. 1=3>2
e. 2>1=3 Intensity = power/area  Emax
Does not depend on frequency/color!
Announcements
• Reading for Friday: 4.5 – 4.6
E(x,t) = Emaxsin(ax-bt)
Light shines
on black tank
full of water.
How much
energy
absorbed?
Intensity = Power = energy/time  (Emax)2  (amplitude of wave)2
area
area
Intensity only depends on the E-field amplitude but not
on the color (or frequency) of the light!
Today (and next class):
Light is a wave! Or is it? (also some E&M refresher)
• HW5 Assigned Today.
• My office hours to discuss exam
(F619):
– Today 2:30-3:30
– Thursday 2-4
– Friday 1-2
• Special guest lecturer next week
Wed and Fri
So we found: light is a wave!
…and things start to make a lot of sense:
• EM waves can are described by Maxwell's theory
• Double slit experiment and other diffraction
phenomena can be explained.
• Explains grating spectrometers etc.
•…
The photoelectric effect (~1900)
The photoelectric effect is a
phenomenon in which electrons
are emitted from matter as a
consequence of absorbing
energy from light.
But then there was this one, little experiment from Mr. Hertz
(~1887) that just couldn’t be explained with EM waves!...
The effect was only observed with UV light, but not so
with red or IR light.  Effect is frequency dependent!?
Actually, some other problems started to surface, such the
lack of an accurate model for the black body radiation…
But Mr. Maxwell told us that the light intensity doesn’t
depend on frequency! (Intensity only depends on |E|2)
Is Mr. Maxwell wrong this time?? He was right last time…
1
Experimental apparatus: PE effect
Experimental apparatus: PE effect
Metal surface
Metal surface
Glass cylinder
Glass cylinder
Vacuum
Vacuum
Adjustable voltage
Current meter
What happens?
A
What happens?
B
A
B
2 ohms
-
10 V
2 ohms
+
-
Two metal plates in vacuum with a voltage between them.
How much current is flowing through the resistor?
A)
0A
B) 0.2 A
C)
5A
D) 10 A
E) infinite current
Potential difference between A and B is 10 V
Without light, no electrons can get across gap.
But if we put an electron close to the surface of plate A it 'feels'
the electric field between the two plates.
The electron accelerates towards the positive plate (B) and picks
up the energy = q(10V) = 1 electron charge x 10 V = 10 eV
Uniform E-field
between plates
A
Constant
force on
electron
constant
acceleration
0V
F=qE
E
+
+
+
+
+
10V
B
10 V
+
Two metal plates in vacuum with a voltage between them.
The potential difference between A and B is:
A) 0 V
B) 5 V
C) 10 V
D) infinite volts
A note about units of energy
Joules are good for macroscopic energy conversions.
But when talking about energy of single electrons,
Joules are inconvenient… (way too big!)
Define new energy unit: The electron-volt (eV)
1eV = kinetic energy gained by an electron when
accelerated through 1 volt of potential difference
1eV ≈ 1.6·10-19 J
0V
- 10V +
0V
F
Current==0.1
0 AA
Current
10Volts
E
path
1V
+
+
+
+
+
1V
2
Example: Different d, same ΔU
Case 1:
0V
distance: d
Case 2:
0V
d/2
1V
+
+
+
+
e-
e-
1V
+
+
+
+
Electron Volt: A convenient unit for energy
Example: KEinit = 2 eV
Electron in both cases initially at rest. What can you say
about the final kinetic energies KE1 and KE2 (case 1 and 2,
respectively) of the electrons just before they hit the right
plate?
Electric Field: E = U / d
A) KE1 > 2 KE2
Force on e:
F = q·E
B) KE1 = 2 KE2
Kinetic energy: ΔKE = F·d
C) 2 KE1 = KE2
Case 1: ΔKE1 = (e·U/d)·d = e·U
D) 2 KE1 < KE2
Case 2: ΔKE2 = (e·U/(d/2))·(d/2) = e·U
E) Something else
 ΔKE1 = ΔKE2
How to put the e- close to plate A?
Shine light on the plate!!
Metal surface
A
Glass cylinder
Electrons get pulled
towards
Vacuum plate B by
F = E·q the electric field
e-
10 V
0V
+
+
+
+
KEx=0.5 = ?
KEfinal = ?
𝐾𝐸𝑓𝑖𝑛𝑎𝑙 = 2𝑒𝑉 + 10𝑒𝑉
What is the kinetic energy half
way between the two plates?
0VElectric field: E = U / d U = 10 V Electric field is the
same everwhere.
So 𝑉 𝑥 = 𝐸 ∙ 𝑑𝑥
E = ½ U / ½d
𝑥
= 10𝑉 ×
𝑑
→
𝐾𝐸
=
2𝑒𝑉
+ 5𝑒𝑉
d
Experimental apparatus: PE effect
Play with color and
intensity. Measure
current I. (I ~ #e-/s)
+
B
+
+
+
+
Adjustable voltage
Current meter
First we could argue that the light heats
up the plate  electrons pop-out
Measure the current!
photoelectric_en.jar
Hot plate.
A few electrons get
enough energy to just
barely “splash” out.
Assume electrons can only move
in the horizontal direction.
Measure the current!
Current
C
0 Voltage
0 Voltage
D
Current
B
0 Voltage
Current
A
Current
What is the current
vs. battery voltage?
0 Voltage
3
What’s happening here?
Each electron that pops out is accelerated and
hits the plate on the right side.
Vacuum tube diode
Current
Current
BUT: # of electrons = constant
Here,
sec
electrons
So current is constant!
are
repelled
by neg.
electrode
not I = V / R !!
0
Battery Voltage
reverse V,
no electrons
flow.
voltage
reverse V,
no electrons
flow.
Vacuum tube diode. Works!
- early electronic device.
photoelectric_en.jar
What do you think does actually happen?
Optical power P
- frequency f
Now: Take out a piece
of paper and draw
the following graphs
with what you expect
will happen.)
Voltage U
Let's do the ‘experiment’!
Play with color and
intensity. Measure
current I. (I ~ #e-/s)
http://phet.colorado.edu
Current I
1. Current vs. Voltage with the lamp on (fixed color,
say UV light, and fixed intensity.)
2. Current vs. Frequency (color) at a fixed intensity
and voltage (right plate is on positive potential)
3. Current vs. Intensity for fixed color (right plate is at
fixed, positive voltage)
photoelectric_en.jar
photoelectric online
Measure the current!
That's what happened:
3. I vs. intensity:
high intensity
low intensity
0
I
2. I vs. f:
Threshold
0
Frequency
I
Threshold
U
0
Intensity
or: Initial KE vs. f:
Initial KE
I
1. Current vs. Voltage:
0
Threshold
Frequency
4