Photoelectric effect

1. The Quantum Theory of Light
and The Photoelectric Effect
Nelson Reference Pages:Nelson Reference Pages:
Page 620 -628Page 620 -628

2. Blackbody Radiation- An Introduction – Do not copy
If a piece of steel is heated with a torch it will initiallyIf a piece of steel is heated with a torch it will initially
not change colour. But, it will give off heat, which ischange colour. But, it will give off heat, which is
invisible because it consists ofinvisible because it consists of infrared radiation.radiation.
Further heating, resulting in a temperature increaseFurther heating, resulting in a temperature increase
will cause the steel to glow red. As the temperaturewill cause the steel to glow red. As the temperature
is further increased, the steel will glowis further increased, the steel will glow yellow andand
further heating will result in the steel glowingfurther heating will result in the steel glowing white..
A further temperature increase will result inA further temperature increase will result in UV
radiation (UV is not visible).radiation (UV is not visible).
As the steel is heated, it initially will emit low frequencyAs the steel is heated, it initially will emit low frequency
(large(large λλ)) electromagnetic radiation (EM) and as theelectromagnetic radiation (EM) and as the
temperature increases so does the frequency of EMtemperature increases so does the frequency of EM
radiation. White light indicates that all frequencies ofradiation. White light indicates that all frequencies of
the visible spectrum are given off.the visible spectrum are given off.

3. Gustav Kirchhoff (1824 - 1887)Gustav Kirchhoff (1824 - 1887) defined thethe
properties of aproperties of a blackbody from his studies on thefrom his studies on the
emission and absorption spectra for gases. Heemission and absorption spectra for gases. He
found that when a gas is heated tofound that when a gas is heated to
incandescence (glows) it will give off specificincandescence (glows) it will give off specific
frequencies of light. And conversely, when whitefrequencies of light. And conversely, when white
light is shone on the same gas, it will absorb thelight is shone on the same gas, it will absorb the
same frequencies as it gave off. Kirchhoffsame frequencies as it gave off. Kirchhoff
proposed thatproposed that all objects will absorb the sameobjects will absorb the same
characteristic frequencies of EM radiation thatcharacteristic frequencies of EM radiation that
they give off. He then defined a blackbody asthey give off. He then defined a blackbody as
the “the “perfect radiator” as it could emit the entire” as it could emit the entire
EM spectrum. Since the blackbodyEM spectrum. Since the blackbody must alsoalso
absorb the entire spectrum, Kirchhoff showedabsorb the entire spectrum, Kirchhoff showed
how this could easily be done using a block withhow this could easily be done using a block with
a small hole and an internal cavity.a small hole and an internal cavity.

4. The imperfect diagram of aThe imperfect diagram of a blackbody radiator isis
given below. It shows a block with a hole, ofgiven below. It shows a block with a hole, of
very small diameter, leading to an inner cavity.very small diameter, leading to an inner cavity.
Any EM radiation entering the cavity through theAny EM radiation entering the cavity through the
hole will not be able to leave. If the inner cavityhole will not be able to leave. If the inner cavity
is continuously heated, the entire EM spectrumis continuously heated, the entire EM spectrum
will be emitted through the tiny hole.will be emitted through the tiny hole.

5. Max Planck’s “Quantum” of Energy - Start
Planck (1858-1947) proposed that the allowed
energy of an oscillator (the e -
had not yet
been discovered) of a specific material could
only exist as a integer (n) multiple of hf, here
h was a constant (Planck’s constant = 6.63 x
10 -34
Js) and f was the frequency. So, E =
nhf represented the amount (quantum) of
energy that could be absorbed or given off.
{This had to do with either absorbing, or
giving off a photon – although this aspect of
EM radiation was not yet known.}

6. The Photoelectric Effect
In 1902 Philipp Lenard, a physicist, performedIn 1902 Philipp Lenard, a physicist, performed
experiments using the setup shown below forexperiments using the setup shown below for
various frequencies of light. Initially, Lenard usedvarious frequencies of light. Initially, Lenard used
UV light and found that a current was present. ThisUV light and found that a current was present. This
indicated that the light was able to freeindicated that the light was able to free e-
whichwhich
could then travel through the vacuum chamber. Ascould then travel through the vacuum chamber. As
thethe intensity of light was increased the ______ alsoof light was increased the ______ also
increased. The ejectedincreased. The ejected e-
were called
photoelectrons.
A
V a c u u m c h a m b e r w i t h
m e t a l l i c p l a t e s a t e a c h e n d

7. Lenard modified his experiment by reversing theLenard modified his experiment by reversing the
terminals of theterminals of the variable voltage power supply. By. By
creating an electric field that opposed the motion ofcreating an electric field that opposed the motion of
the electrons, Lenard was able to determine Ek ofthe electrons, Lenard was able to determine Ek of
the photoelectrons by adjusting the voltage until thethe photoelectrons by adjusting the voltage until the
current completely stopped. This was called thecurrent completely stopped. This was called the
stopping PD (depended on the light frequency and
type of metal plate used)..
A
V a c u u m c h a m b e r w i t h
m e t a l l i c p l a t e s a t e a c h e n d
- +

8. Lenard also used a prism to breakup whiteLenard also used a prism to breakup white
light and then used different colours of light.light and then used different colours of light.
He found that the stopping PD was greater forHe found that the stopping PD was greater for
higher frequency light (smallerfrequency light (smaller λλ) and that) and that
higher intensity light of a lower frequency
could not compensate for a stopping
potential set for a higher frequency light..
TheThe conclusion was that higher frequency lightwas that higher frequency light
would produce photoelectrons with greaterwould produce photoelectrons with greater
Ek and light intensity wasEk and light intensity was not a factor ina factor in
achieving theachieving the maximum Ek for the photo efor the photo e--
..

9. Einstein and the Photoelectric Effect (PE)
In 1905 Einstein showed the relationship between
Planck’s quantum of energy and the PE. Einstein
proposed the following:
1. Light must be emitted and absorbed as a quantum
of energy. These quanta, or packets of energy,
were later called photons.
2. Planck’s unit of energy, hf, was the energy of a
_________ .
3. When a photon strikes a metal surface, all of the
energy of the photon is absorbed by one e-
in
one event.
4. Since the energy of a photon is related to its
frequency, a higher frequency photon will transfer
more energy to an e-
then a lower frequency
photon.

10. Einstein also proposed
answers to other issues
related to the PE:
 Some photo e-
had more
Ek than others. This was
related to how “deep” the
e-
was buried. Those e-
buried deeper required
more energy to be freed
from the metal and thus
had less Ek energy. The
e-
closest to the surface
were the easiest to
remove and would have
max Ek. Einstein named
the minimum amount of
energy to remove an e-
as the work function
(W). The value of
W depends on the
type of metal
(values pg 621).

11. The maximum Ek for a photoThe maximum Ek for a photo e-
is:
Ek = hf – W, {recall 1.0 eV = 1.6 x 10recall 1.0 eV = 1.6 x 10-19-19
JJ}
Einstein did receive the Nobel Prize in 1921 for his
contributions to the understanding of the
photoelectric effect.
The following video summarizes the PE.The following video summarizes the PE.
 http://www.youtube.com/watch?v=0qKrOF-gJZ4http://www.youtube.com/watch?v=0qKrOF-gJZ4
The next video shows the effect of light color on currentThe next video shows the effect of light color on current
 http://www.youtube.com/watch?v=kcSYV8bJox8
Threshold Frequency (f0)
Using Ek = hf – W, the lowest frequency to free an e-
is called the threshold frequency and does depend
on the type of metal. This freed photo e-
will have
no Ek and will be re-absorbed by the metal.

12. Robert Millikan and the
Photoelectric Effect
Millikan performed experiments
using different metals to find
the Ek of the photoelectrons.
The vertical axis is a measure
of Ek and the horizontal axis
indicates the frequency of
light used. The intersection of
each line with the frequency
axis represents the _____
_____ . If the line is continued
below the f axis it would
intersect the Ek axis at the
value of the ______ _______ .
{See pg 623}
Which metal
requires the
most energy to
remove an e-
?
Ans. Na
K N a
f0 ,K f0 ,N a f0
E k v s f
E k

13. Momentum of a Photon
Although a photon canAlthough a photon can
not have a rest mass,not have a rest mass,
it can be shownit can be shown
mathematically that amathematically that a
photon does possessphoton does possess
momentum. Themomentum. The
equation on the rightequation on the right
equates Einstein’sequates Einstein’s
famous E = mcfamous E = mc22
to theto the
energy of a photon.energy of a photon.
 E = mcE = mc22
and E = hfand E = hf
 So, hf = mcSo, hf = mc22
 But p = mv = mcBut p = mv = mc
 So, hf = p cSo, hf = p c
 p = hf/cp = hf/c
 ButBut λ = c/fλ = c/f
 So,So, p = h/p = h/λλ

14. Practice Questions
 Nelson Textbook
 Page 624 #1, 2 Page 626 #1-3Page 624 #1, 2 Page 626 #1-3
 Page 631#1 - 6Page 631#1 - 6
From McGraw-Hill Textbook
The eThe e--
emitted from a surface illuminated byemitted from a surface illuminated by
light of λ = 460 nm has a max speed oflight of λ = 460 nm has a max speed of
4.2 x 104.2 x 10 55
m/s. Calculate W (in eV) of them/s. Calculate W (in eV) of the
surface material. (Ans 2.2 eV)surface material. (Ans 2.2 eV)
 Workbook: Page 54 #1-5