AP Physics – Quantum Mechanics
Problems with behavior of electrons
Black Body Problem
Higher the temp, greater the freq, shorter the wavelength
Intensity
4000 K
3000 K
2000 K
Wavelength (  m)
Total radiation -- area under curve
Increases with T
Classical physics could not explain Curve
1
Intensity
Classical
Theory
Planck’s Theory
Wavelength
Classical mechanics works -- long wavelengths
Classical mechanics predicted E would approach infinity as wavelength approached zero
E  hf
Max Planck
h  6.626 x 1034 J  s
h  4.14  1015 eV  s

What energy is carried by a quantum (photon) of electromagnetic radiation that has a
frequency of 1.55 x 1017 Hz?
E  hf
1

 6.63 x 1034 J s 1.55 x 1017 
s

E  10.3 x 1017 J

1.03 x 1016 J
Energy as function of wavelength:
2
cf
f 
c
E  hf

E
E
hc

hc

Photoelectric Effect
Light incident on metals – electrons get emitted
  Work Function
Energy that binds electron to metal
Depends on metal
KMax  hf  
Photon’s energy less work function
Table 27.1 pg 890
Look at wavelength
c
c f
K Max 
hc


 f

Equation on AP test
Values for hc terms:
hc  1.99 x 1025 J  m
hc  1.24 x 103 eV  nm
These values are on AP test
3

What is the velocity of a photoelectron that has been liberated from a zinc metal
surface by a photon that has a wavelength of 275 nm?
work function for zinc is 4.31 eV
K Max 
hc

1.24 x 103 eV  nm

 4.31 eV
275 nm

K Max  0.00451 x 103 eV  4.31 eV
KMax  4.51 eV  4.31 eV  0.200 eV
 1.6 x 1019 J 
19
K Max  0.200 eV 
  0.32 x 10 J
1 eV


1
K  mv 2
2
v
2K
m
2

20 kg  m
2  3.2 x 10


s 2 

v

9.11 x 1031 kg
m2
v  7.025 x 10
s2
10
 3.2 x 1020 J

m2
0.7025 x 10 2
s
2.65 x 105
11
m
s
Photoelectric Lab Setup:
4
Emitter  Plate at neg potential
Light strikes emitter
C
E
e
A
V
Variable power
supply
Electrons emitted
Electrons travel to Collector
Vary wavelength:
intensity constant,
Get current curve
Current
0  photo-electric threshold wavelength
Current Depends on wavelength
Increase V -- max current
Even at zero potential get some current
0
Wavelength
Make Collector negative
5
-- Collector Repels electrons
-- Current drops to low value
High Intensity
Current
Low Intensity
  VS
Applied Voltage
V less than or equal to Vs no electrons reach C
All electrons get turned away
Vs  Stopping potential
current is zero
no electrons reach Collector
Stopping potential independent of light intensity
Max K is energy gained in field (PE:
U E  qV
SO
KMax  qVs
KMax independent of intensity!
Einstein explained in 1905 (Noble Prize)
Electrons bound to metal
Photon gives all its energy to single electron
Electron gains energy and is liberated
6
Energy gained exceeds the energy that binds it to the metal
graph of frequency VS. kinetic energy
Has straight line.
KE
fC

cutoff
frequency
Slope of
graph is h
Cutoff
f
fc
Frequency -minimum
frequency that
will generate photoelectrons
minimum frequency is when kinetic energy = zero.
Set KMax = 0
K Max  hf  
fC 
hf  
f 

h

h
Cutoff Wavelength
Set KMax = 0
7
–photon E can just overcome work function
K Max 
0
hc

hc


hc



hc


C 
hc

be prepared to derive these equations on AP test

What is the cutoff wavelength for a copper metal surface?
C 
hc

1.24 x 103 eV  nm

4.70 eV
 0.264 x 103 nm 
264 nm
wavelength is smaller than visible light
Finding Work Function:
- solve for 
Set KMax = 0
Use cutoff wavelength
K Max 
hc
C

0
hc
C



hc
C
hc
C
Also can find as function of cutoff frequency.
KMax  hf  
0  hf  
  hf
  hfC
8

500.0 nm light is incident on a metal surface. The stopping potential is found to be
0.440 V. (a) Find the work function for this material and (b) the longest wavelength
that will eject electrons from the metal.
(a) work function:
K Max 
hc
qVs 
hc



K Max  qVs


hc

 qVs
1.24 x 103 eV  nm

 e  0.440 V   0.00248 x 103 eV  0.440 eV
500 nm
  2.48 eV  0.440 eV

2.04 eV
(b) longest wavelength
C 
hc

1.24 x 103 eV  nm

204 eV
 0.00608 x 103 nm 
608 nm
Classical Mechanics or wave theory cannot explain:



No electrons emitted if light freq falls below cutoff freq, fC
Max K independent of light intensity
Electrons emitted almost instantaneously
 KEMax increases with increasing freq as it is function of hf
 Happens so fast because it is a one to one photon/electron deal
9
AP Test:
A sodium photoelectric surface with work function 2.3 eV is illuminated
by electromagnetic radiation and emits electrons. The electrons travel
toward a negatively charged cathode and complete the circuit shown
below. The potential difference supplied by the power supply is
increased, and when it reaches 4.5 V, no electrons reach the cathode.
(a) For the electrons emitted from the
sodium surface, calculate the following.
i. The maximum kinetic energy.
KMax  qVs


K Max  1.6 x 1019 C  4.5 V  
7.2 x 1019 J
ii. The speed at this maximum kinetic
energy.
1
K  mv 2
2
v
2K
m
2


19 kg  m
2  7.2 x 10
2
s



31
9.11 x 10 Kg

1.26 x 106
m
s
(b) Calculate the wavelength of the radiation that is incident on the sodium
surface.
K Max 

hc


hc
 KEMax   
K Max   
hc

  K Max     hc  
1.24 x 103 eV  nm

 4.5 eV  2.3 eV 
hc
 K Max   
 0.182 x 103 nm  182 nm
(c) Calculate the minimum frequency of light that will cause photoemission from
this sodium surface.
KMax  hf  
KMax  0 for minimum Freq
0  hf  
hf  
f 

h
10
f 
2.3 eV
4.14 x 1015 eV  s
 0.555 x 1015 Hz
 5.55 x 1014 Hz
Stopping Potential versus frequency
Stopping potential is function of frequency
Higher frequency  higher VS
Vs
f
fc
intercept on x axis is stopping potential for cutoff frequency.
At fC - total
E = 
E    hfC
E  qVS
E  eVS
This energy = energy of photon, so:
hf  eVs
solve for h/e
h Vs

e
f
11
h/e is slope of the graph
AP Test Item:

In a photoelectric experiment, light is incident on a metal surface. Electrons are
ejected from the surface, producing a current in a circuit. A reverse potential is
applied in the circuit and adjusted until the current drops to zero. That potential at
which the current drops to zero is called the stopping potential. The data obtained for
a range of frequencies are graphed below.
a. For a frequency of light that has a stopping potential of 3.0 volts, what is the
maximum kinetic energy of the ejected photoelectrons?
K Max  qVs
 e  3.0 V  
3.0 eV
b. From the graph and the value of the electron charge, determine an experimental value
for Planck's constant.
Slope of graph is
h
2V  0V

e 10  51014 Hz
h
e
 h
1.6 x 1019 C  2 V  0 V 
10  51014 Hz
12
3.2 x 1019 J
h
1
5 x 1014
s
h V2  V1

e f 2  f1
h
e2 V 
1
5 x 1014
s

 0.64 x 1033 J  s 
6.4 x 1034 J  s
2V  0V
10  51014 Hz
e2 V 
1
5 x 1014
s
h
 0.40 x 1014 eV  s 
or
4.0 x 1015 eV  s
c. From the graph, determine the work function for the metal.
1

 4.14  1015 eV  s  5.0 x 1014   20.7  101 eV  2.07 eV g
s

h
raph of straight line
y  mx  b
Vs  f  b
eVs  hf  eb
e
K Max  hf  eb
But eVs is KMax
  hfC
so y intercept is work function  times e
This is 2.0 eV
d. On the axes above, draw the expected graph for a different metal surface with
a threshold frequency of 6.0 x 1014 hertz.
13
Photons and Power:
Given power of light source and wavelength
Can calculate number of photons per second.

An LED produces 685 nm light at a power rating of 1.25 x 10-7 W. Find the number
of photons emitted per second.
P
E
t
E  Pt
let t = 1 second
 1.25 x 107
solve for E
J
1 s   1.25 x 107 J
s
Energy of one photon:
E
hc

1.24 x 103 eV  nm

685 nm
1.25 x 10
7

1 eV
J 
19
 1.6 x 10 J

 0.00181 x 103 eV
 1.81 eV
  1 photon 
12

  0.432 x 10 photon
  1.81 eV 
4.32 x 1011 photon
14