A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
©
2007
• Discuss the meaning of quantum physics
and Planck’s constant for the description of
matter in terms of waves or particles.
• Demonstrate your understanding of the
photoelectric effect, the stopping potential,
and the deBroglie wavelength.
• Explain and solve problems similar to those
presented in this unit.
In his studies of black-body radiation, Maxwell
Planck discovered that electromagnetic energy is
emitted or absorbed in discrete quantities.
Planck’s
Equation:
E = hf
(h = 6.626 x 10-34 J s)
Apparently, light consists of
tiny bundles of energy called
photons, each having a welldefined quantum of energy.
Photon
E = hf
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
First we find f from wave equation: c = fl
f 
c
l
;
E  hf 
hc
l
(6.626 x 1034 J  s)(3 x 108 m/s)
E
-9
555 x 10 m
E = 3.58 x 10-19 J
Or
E = 2.24 eV
Since 1 eV = 1.60 x 10-19 J
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 . . .
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.
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.
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
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
-
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
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
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
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:
Recall that the formula for the relativistic total
energy was given by:
Total Energy, E
E  (m0c 2 )  p 2c 2
For a particle with zero
momentum p = 0:
A light photon has
mo = 0, but it does
have momentum p:
E = m oc 2
E = pc
We know that light behaves as both a wave and
a particle. The rest mass of a photon is zero, and
its wavelength can be found from momentum.
E  pc 
hc
l
Wavelength
of a photon:
h
l
p
All objects, not just EM waves, have wavelengths
which can be found from their momentum
de Broglie
Wavelength:
h
l
mv
In working with particles of momentum p = mv,
it is often necessary to find the momentum from
the given kinetic energy K. Recall the formulas:
K = ½mv2 ;
Multiply first
Equation by m:
p = mv
mK = ½m2v2 = ½p2
Momentum from K:
p  2mK
 1.6 x 10-19 J 
-17
K  90 eV 

1.44
x
10
J

 1 eV 
Next, we find momentum
from the kinetic energy:
p  2mK
p  2(9.1 x 10-31kg)(1.44 x 10-17 J)
p = 5.12 x 10-24 kg m/s
h
6.23 x 10-34 J
l 
-24
p 5.12 x 10 kg m/s
e- 90 eV
h
h
l 
p mv
l = 0.122 nm
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
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
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
Quantum physics works for waves or particles:
For a particle with zero
momentum p = 0:
A light photon has
mo = 0, but it does
have momentum p:
Wavelength
l
of a photon:
h

p
E = m oc 2
E = pc
h
de Broglie
l
Wavelength:
mv