Download kinetic energy of photoelectrons (eV)

“The temperature of a lava flow can
be approximated by merely
observing its colour. The result
agrees nicely with the measured
temperatures of lava flows at about
1,000 to 1,200 °C.”
• In the late 19th Century,
physicists began to study
light emitted from hot
solids and gases
• They studied nonreflecting solids that
appeared black at room
temperature
Why do black objects appear black?
•When a solid object is heated, it emits EMR with a
range of frequencies. (The atoms are vibrating –
as predicted by Maxwell’s equations.)
• It followed that objects at higher temp’s should
emit EMR with higher frequencies.
frequency of wave animation
• Classical theory predicted that the intensity of the radiation
would increase as the square of the frequency but this did
not fit with observations
It’s an
ultraviolet
catastrophe!
As the temp. increased:
 the max brightness (intensity) increased and more
EMR was emitted as shorter wavelengths (higher freq.)
However, the
frequencies
reached a peak,
then dropped.
Black Body Animation
•This contradiction was called the
ultraviolet catastrophe
•Max Planck developed
a theory that proposed
that the vibrating atoms
could only have, emit,
or absorb certain
discrete amounts of
energy (or whole
number multiples)
•He called the bundles
quanta (singular:
quantum)
•Planck’s theory was revolutionary because it proposed
that an object cannot vibrate with any amount of energy
– only specific amounts.
•Whereas, classical theory said energy could be sent
out in continuous streams of waves
 A quantum of light is
known now as a photon
… may be thought of as a wave packet
kinda like:
The energy of a quantum (photon) of light was proportional
to its frequency: Planck's Law
E  hf
• Planck’s
equation fit all
observations of
blackbody
radiation
• Since c = f, Planck’s Law can be written
E
hc

Example 1
Determine the energy of a photon of light (in J) with a
frequency of 2.56 x 1014 Hz.
E = 1.70 x 10-19 J
• The energy of a photon is very small, a more convenient
unit is the “electron volt” (eV)
 the amount of energy an electron gains or loses as
it moves through a potential difference of 1 volt
-19
10
1 eV = 1.60 x
J
-15
h = 4.14 x 10 eVs
Example 2
Determine the energy of a photon in eV, of light with a
frequency of 4.88 x 1014 Hz.
E = 2.02 eV
Example 3
Determine the energy in eV of a photon emitted by a
laser in a retail scanner (λ = 632 nm).
E = 1.97 eV
Example 4
The same retail scanner expends 1.0 mW of power.
How many photons are emitted per second?
E = 3.17 x1015 photons / second
• Hertz noticed that certain metals
lost negative charges when
exposed to ultraviolet light
• The term PHOTOELECTRIC
EFFECT was given to this
phenomenon and the electrons
given off were called
photoelectrons
• Many experiments were
done to measure the effects of
the incident radiation on the
energy and number of the
photoelectrons
animation
 As the voltage increases, the current through the tube
decreases to zero, this is known as the stopping voltage and
is related to the maximum kinetic energy of the
photoelectrons.
Must be in Joules!
Ek max  qVstop
Example #1
Photoelectrons emitted from a piece of selenium have a
maximum kinetic energy of 7.2 eV. Calculate the stopping
voltage of the photoelectrons.
Vstop = 7.2 V
 Experiments on lots of different metals and different
frequencies of incident light gave results which could not
be completely explained by classical physics:
1) Electrons are emitted from the metal only if the
incident frequency is above a certain threshold
frequency (fo).
2) Intensity (brightness of the light) had no effect on fo.
No matter how bright the light, if it is below fo, no
electrons are emitted.
3) Each metal has its own value of fo.
4) As the frequency of the incident radiation increased, the
kinetic energy of the photoelectrons increased
5) The photoelectrons are emitted immediately (classical
physics predicted time delays of weeks for very faint
light)
6) The brightness of the light increased the amount of
photoelectrons, but did not change the kinetic energy
• Einstein proposed that
each photon gave all of its
energy to just one electron.
•The amount of energy was
given by Planck’s equation
• The electron required a
certain amount of energy toEnergy of
be freed from its atom.
incident
 called the work
photon
function
E
  Eout
hf  Ekmax  W
in
Energy of
Work
photoelectron function
W  hf o
• Electrons emitted from the
surface of a metal would
have the maximum amount
of kinetic energy limited by
the work function
You Loser!
kinetic energy of photoelectrons (eV)
Kinetic energy of photoelectrons as a
function of incident frequency
Metal
1
Metal
2
Metal
3
incident frequency (Hz)
• The slope of the lines is Planck’s constant,
x-intercept is fo, the y-intercept is the work
function of the metal
 Millikan used the photoelectric effect to
accurately determine Planck’s constant and
to verify Einstein’s work
Practice Problem #1
Determine the work function
of a metal which emits
photoelectrons only when the
wavelength of the incident
light is equal to or greater
than 500 nm.
Get it? … it’s a
“work
function!”
W = 2.48 eV
Practice Problem #2
Light with a frequency of 7.00 x1015 Hz strikes a crystal inside
of a photovoltaic cell which has a work function of 1.50 eV.
Determine the kinetic energy of the photoelectrons.
Ekmax = 27.5 eV
Practice Problem #3
Determine the maximum speed of the photoelectrons in
the last example.
Remember:
If it has mass, it’s
moving slower than
3.00 x 108 m/s!
v = 3.11 x106 m/s