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Plan for Today (AP Physics 2)
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• Notes/Lecture on Photoelectric Effect
Background
• In the late 1800s, scientists discovered
that when light shines on certain metallic
surfaces, electrons are emitted
• This is called the photoelectric effect
• Emitted electrons are photoelectrons
Big Idea with Set Up
• If there is sufficient energy in the light,
the electrons can be freed
• The electrons would then go from the
emitter to the collector (well, some
would)
• Electrons would be freed only if there is
enough energy to free them
Energy at this level
• Photons = packets of energy
• Einstein extended Planck’s idea of energy
quantization to electromagnetic waves
• Photon is a tiny packet of light energy
• Made when a quantized oscillator jumps
from a higher energy state (E = nhf) to a
lower state (E = (n-1)hf).
• Because of conservation of energy, the
decrease in energy hf must be the photon’s
energy
Energy of a Photon
• E = hf
• h is Planck’s constant
• F is frequency of the light
Energy in Electron-volts
Photon energies are so small that the energy is
better expressed in terms of the electron-volt.
One electron-volt (eV) is the energy of an
electron when accelerated through a potential
difference of one volt.
1 eV = 1.60 x 10-19 J
1 keV = 1.6 x 10-16 J
1 MeV = 1.6 x 10-13 J
Example 1: What is the energy of a photon of
yellow-green light (l = 555 nm)?
First we find f from wave equation: c = fl
f 
c
l
;
E  hf 
34
hc
l
(6.626 x 10 J  s)(3 x 10 m/s)
E
555 x 10-9 m
E = 3.58 x 10-19 J
8
Or
E = 2.24 eV
Since 1 eV = 1.60 x 10-19 J
Useful Energy Conversion
Since light is often described by its wavelength in
nanometers (nm) and its energy E is given in eV, a
conversion formula is useful. (1 nm = 1 x 10-9 m)
E (in Joules) 
hc
l
; 1 eV  1.60 x 10-19 J
hc(1 x 109 nm/m)
E (in eV) 
(1.6 x 10-19 J/eV)l
If l is in nm, the energy in eV is found from:
E
1240
l
Verify the answer
in Example 1 . . .
Back to the Photoelectric
Effect
Photoelectric Set Up
• Photocell (evacuated glass tube) contains a
metal plate (E) connected to the negative side
of a power supply. Another plate C is connected
to the positive end of the power supply.
• When light shines on E (of a particular
wavelength), we have a current!
• The current is from photoelectrons emitted from
the negative plate (emitter) and collected at the
positive plate (collector)
Diagram of Set Up
Photoelectric Effect and
Potential Difference
• At a large potential difference, current
reaches a maximum
• Current increases as incident light
intensity increases
Now what if we flip the
battery
• This way the collector plate is negative and
the emitter plate is positive
• Then the current drops to a lower value
because most of the emitted photoelectrons
are repelled by the collector plate
• Only electrons with enough KE to overcome
the repulsion reach the collector plate
Battery Flipped Continued
• When the potential difference is more
negative than Vs, no electrons reach the
collector and current is 0
• This point Vs is called the stopping
potential
• Stopping potential is independent of
radiation intensity
Stopping Potential
Maximum Kinetic Energy of
Photoelectrons
• KEmax = e * Stopping Potential
• e is the charge of an electron
Problems with Classical Physics
(Why the Photoelectric Effect)
• No electrons emitted if light frequency is
below a cutoff frequency (for a given
material)
– We would expect photoelectric effect at any
frequency with enough intensity
• Maximum KE is independent of light intensity
– We would expect higher intensity means more
energy means more KE in photoelectrons
More Problems
• Max KE of photoelectrons increases with
increases frequency
– We wouldn’t expect there to be a
relationship
• Electrons emitted almost instantaneously
– We would expect a bit of a delay as
photoelectrons absorb energy
The Photo-Electric Effect
Incident light
Cathode
Anode
A
C
-
+
Ammeter
A
When light shines on
the cathode C of a
photocell, electrons are
ejected from A and
attracted by the positive
potential due to battery.
There is a certain threshold energy, called the
work function W (
), that must be overcome
before any electrons can be emitted.
Photons must give some of their energy here first
Photoelectric Effect Equation
• KEmax = hf – W
• Work function is the minimum energy
with which an electron is bound in the
metal
Cutoff Wavelength
• Graph of frequency vs. KEmax gives us a
linear relationship
• X intercept (horizontal frequency axis)
gives us the cutoff frequency where no
photoelectrons are emitted
Cutoff Wavelength
• Waves greater than the cutoff
wavelength do not result in emission of
photoelectrons
Photo-Electric Equation
Incident light
Cathode
Anode
A
C
-
+
Ammeter
A
E
hc
l
 W  12 mv 2
hc
Threshold
W
wavelength lo
l0
The conservation of energy demands that the
energy of the incoming light hc/l be equal to the
work function W of the surface plus the kinetic
energy ½mv2 of the emitted electrons.
Example 2: The threshold wavelength of light
for a given surface is 600 nm. What is the
kinetic energy of emitted electrons if light of
wavelength 450 nm shines on the metal?
hc
l = 600 nm
W  K
l
hc hc

K
l l0
hc
A
hc
1240
1240
K



;
l l0 450 nm 600 nm
K = 0.690 eV
Or
K = 2.76 eV – 2.07 eV
K = 1.10 x 10-19 J
Stopping Potential
A potentiometer is used
to vary to the voltage V
between the electrodes.
The stopping potential
is that voltage Vo that
just stops the emission
of electrons, and thus
equals their original K.E.
Photoelectric equation:
E  hf  W  eV0
Incident light
Cathode
Anode
V
A
+
-
Potentiometer
Kmax = eVo
W
h
V0    f 
e
e
Slope of a Straight Line (Review)
The general equation for
a straight line is:
y = mx + b
The x-intercept xo occurs
when line crosses x axis
or when y = 0.
The slope of the line is
the rise over the run:
The slope of a line:
y
Slope
y
x
xo
x
y
Slope 
m
x
Finding Planck’s Constant, h
Using the apparatus on the previous slide, we
determine the stopping potential for a number
of incident light frequencies, then plot a graph.
W
h
V0    f 
e
e
Finding h constant
V
h
Slope 
e
Note that the x-intercept fo
is the threshold frequency.
Stopping
potential
Slope
fo
y
x
Frequency
Example 3: In an experiment to determine
Planck’s constant, a plot of stopping potential
versus frequency is made. The slope of the
curve is 4.13 x 10-15 V/Hz. What is Planck’s
constant?
Stopping
potential
V
Slope
fo
y
x
Frequency
W
h
V0    f 
e
e
h
-15
Slope   4.13 x 10 V/Hz
e
h = e(slope) = (1.6 x 10-19C)(4.13 x 10-15 V/Hz)
Experimental Planck’s h = 6.61 x 10-34 J/Hz
Example 4: The threshold frequency for a given
surface is 1.09 x 1015 Hz. What is the stopping
potential for incident light whose photon energy
is 8.48 x 10-19 J?
Photoelectric Equation:
E  hf  W  eV0
Incident light
Cathode
Anode
V
eV0  E  W ; W  hf 0
A
+
-
W = (6.63 x 10-34 Js)(1.09 x 1015 Hz) =7.20 x 10-19 J
eV0  8.48 x 10 J  7.20 x 10 J  1.28 x 10 J
-19
1.28 x 10-19 J
V0 
1.6 x 10-19 J
-19
-19
Stopping
Vo = 0.800 V
potential:
Summary
Apparently, light consists of
tiny bundles of energy called
photons, each having a welldefined quantum of energy.
Planck’s
Equation:
E = hf
Photon
E = hf
(h = 6.626 x 10-34 J s)
The Electron-volt:
1 eV = 1.60 x 10-19 J
1 keV = 1.6 x 10-16 J
1 MeV = 1.6 x 10-13 J
Summary (Cont.)
Incident light
Cathode
C
-
Anode
A
+
Ammeter
A
E
hc
l
 W  12 mv 2
hc
Threshold
W
wavelength lo
l0
If l is in nm, the energy in eV is found from:
Wavelength in nm;
Energy in eV
E
1240
l
Summary (Cont.)
Planck’s Experiment:
Incident light
Cathode
Anode
Stopping
potential
V
Slope
fo
y
x
Frequency
V
A
+
-
Potentiometer
Kmax = eVo
W
h
V0    f 
e
e
h
Slope 
e