REVISION
PHOTOELECTRIC EFFECT
PHOTOELECTRIC EFFECT
the process whereby electrons are ejected from a metal
surface when light of suitable frequency is incident on
that surface..
IT PROVES
... the particle nature of light
THRESHOLD FREQUENCY
fo, as the minimum frequency of light needed to emit
electrons from a certain metal surface.
WORK FUNCTION
Wo, the minimum energy that an electron in the metal
needs to be emitted from the metal surface
hfo
WEo  hf
E  Energy (Joule)
h  Planck's constant (6.63 1034 )
f  frequency (Hertz)
E  Wo  Ek
E  hf
1 2
Ek  mv
2
Different metals have different work functions, because
they have different ionization energies
Ionization energy is the energy required to remove valence
electrons from outer orbitals
3 SCENARIOS
E  hf
Photo
electrons are
NOT emitted
E  Wo
Photo
electrons are
emitted
E  Wo
Photo
electrons are
emitted and
move away
using kinetic
energy
E  Wo
E  Wo  Ek
A learner, who wants to demonstrate the photoelectric effect,
uses a zinc plate fitted to the top of a gold leaf electroscope.
The work function of zinc is 6.9  10 19 J
a
Define the concept work function
b
Calculate the maximum wavelength of light that will emit
electrons from the zinc plate
2.88  107 m
The electroscope is negatively charged and UV light is shone on it.
One of the wavelengths of the light is 260nm.
c
Calculate the kinetic energy of a photoelectron emitted from
the surface of the zinc plate
7.5  1020 J
When the learner tries to replicate the experiment with a positively
charged electroscope, he finds that the UV light has no noticeable effect
d
Explain this observation
a  Wo  6.9  1019 J
 ?
h  6.63  1034
E  hf
Wo  hf o
6.9  1019   6.63  1034  f o
c  3  108
b    260  109
Ek  ?
Wo  6.9  1019
c f
3  108  1.04  1015  
  2.88  107 m
f o  1.04  1015 Hz
c f
3  108  f  260  109 
f  1.15  1015 Hz
E  Wo  Ek
7.65  1019  6.9  1019  Ek
Ek  7.5  1020 J
h  6.63  1034
c  3  108
E  hf
  6.63  1034 1.15  1015 
 7.65  1019 J
c  The positive zinc plate attracts the electrons and prevents the emision of photo electrons
INTENSITY & FREQUENCY
EXPERIMENT
• http://phet.colorado.edu/en/simulation/photoelectric
INTENSITY & FREQUENCY
• To increase the CURRENT the rate of flow of current
must be increased
Q
I
t
FREQUENCY
• Frequency depends on the wavelength of light.
• Frequency influences the amount of energy of
the photons.
• To increase the CURRENT
• The photons require more energy in order to
increase the electrons kinetic energy
• Therefore we INCREASE the FREQUENCY
• (or DECREASE the WAVELENGTH)
INTENSITY
• Intensity depends on the number of light sources
• Intensity influences the number of photons that
are released
• To increase the CURRENT
• More electrons need to be release per time
• Which means that more photons need to reach
the surface
• Therefore we INCREASE the INTENSITY
• Increasing the INTENSITY does NOT change the
kinetic energy of the photo-electrons
• (because the energy of the photons didn’t
change)
• So if there AREN’T any photo-electrons released
at a specific frequency, it will not help to increase
the intensity, because the photons will still lack
sufficient energy
• To change the speed of the photo-electrons, the
frequency (energy) of the photons needs to
change
ABSORPTION & EMISSION
SPECTRA
ABSORPTION & EMISSION
SPECTRA
• An atomic absorption spectrum is formed when
certain frequencies of electromagnetic radiation
that passes through a medium, e.g. a cold gas, is
absorbed
• An atomic emission spectrum is formed when
certain frequencies of electromagnetic radiation are
emitted due to an atom's electrons making a
transition from a high-energy state to a lower
energy state.
HYDROGEN
ABSORPTION vs EMISSION
An automatic camera has an inbuilt light sensor.
When light reaches the sensor, it falls on a metal
object that emits electrons and produces a current.
a
What phenomenon is decribed above?
A metal plate is irradiated with electromagnetic
radiation with a wavelength of 200 nm
The work function of the metal is 7.57  10 19 J .
b
Show, by calculation, that photoelectrons will
be emitted.
E>Wo
The intensity of the radiation is increased, while the
wavelength is kept constant.
State and explain the effect this change will have on
c
the energy of the emitted photoelectrons
d
the number of photoelectrons emitted
same
increase
a
Photo electric effect
b    220 109
Wo  7.57 1019 J
E ?
c f
3 108  f  220 109 
f  1.363 1015 Hz
E  hf
  6.63 10 34 1.363 1015 
 9.04 1019 J
c  3 108
E  Wo
h  6.63 1034
Energy of the photons is more
than the work function,  photo
electrons are emitted
c
The energy of the emitted electrons stays the SAMEm because the frequency did not change
d  The number of photo electrons INCREASE
An increase in intensity increases the amount of photons that can transfer energy to the
electrons