Transcript 6.4 - Hockerill Students

6. Atomic and
Nuclear Physics
Chapter 6.4 Interactions of matter
with energy
The Photoelectric effect
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Heinrich Hertz first observed this photoelectric effect in 1887.
This, too, was one of those handful of phenomena that
Classical Physics could not explain.
Hertz had observed that, under the right conditions, when light
is shined on a metal, electrons are released.
The photoelectric
effect consists on the
emission of electrons
from a metallic surface
by absorption of light
(electromagnetic
radiation).
Photoelectrons
Photosurface
The Photoelectric effect

An apparatus to investigate the photoelectric effect was set
by Millikan and it allowed him to determine the charge of the
electron.

When light falls on the
surface, the electrons are
removed from the metal’s
atoms and move towards
the positive cathode
completing the circuit and
thus creating a current.
http://phet.colorado.edu/simulations/sims.php?sim=Photoelectric_Effect
The Photoelectric effect

Whether the photoelectric effect occurs or not depends only on:

The nature of the photosurface
The
frequency of the radiation
The Photoelectric effect
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When the intensity of the light source increases so does the
current.
Current and intensity are directly proportional
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A high current can be due to:
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Electrons with high speed
Large number of electrons being emitted
To determine what exactly happens we need to be able to
determine the energy of the emitted electrons.
This is done by connecting a battery between the
photosurface and the collecting plate.
The Photoelectric effect
light
collecting plate
electron
photosurface
evacuated tube
G
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When the battery supplies a p.d. the charge of the collecting
plate will be negative.
This means that the negatively charged electrons can be
stopped if a sufficiently negative p.d. is applied to the
electrodes.
The Photoelectric effect
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The electrons leave the photosurface with
a certain amount of kinetic energy – EK.
To
stop the electrons we must supply a potential difference
(called stopping voltage) so that:
eVs = EK
The Photoelectric effect
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The stopping voltage stays the same no matter what the
intensity of the light source is.
This means that:
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The intensity of light affects the number of electrons emitted but not
their energy
The energy of the electrons depends on the nature of light: the larger
the frequency, the larger the energy of the emitted electrons and thus
the larger the stopping voltage
Critical of threshold frequency
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The two graphs represent the EK of
the electrons versus frequency

These graphs tell us that:

there is a minimum frequency fc, called
critical or threshold frequency, such
that no electrons are emitted.

if the frequency of the light source is less
than fc then the photoelectric effect does
not occur

the threshold frequency only depends on
the nature of the photosurface
The Photoelectric effect - Observations
1.
The intensity of the incident light does not affect the
energy of the emitted electrons (only their number)
2.
The electron energy depends on the frequency of the
incident light, and there is a certain minimum frequency
below which no electrons are emitted.
3.
Electrons are emitted with no time delay – instantaneous
effect.
The Photoelectric effect
Problem:
According to Classical Physics, the electron should be
able to absorb the energy from light waves and
accumulate it until it is enough to be emitted.
Solution:
Einstein suggested that light could be considered particles
of light, photons, packets of energy and momentum or
quanta. The energy of such quantum is give by:
E=hf
where: f is the frequency of the e-m radiation
h =6.63x10-34J (constant known as Planck constant)
The Photoelectric effect
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When a photon hits a photosurface, an electron will absorb
that energy.
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However, part of that energy will be used to pull the
electron from the nucleus.
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That energy is called the work function and represented
by Ф.
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The remaining energy will be the kinetic energy of the free
electron.
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So,
Ek = hf - Ф
The Photoelectric effect
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Recalling that
EK = eVs
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So,
eVs = hf – Ф
that is:
h

Vs  f 
e
e
The Photoelectric effect
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When a photon hits a photosurface, 3 things can happen:
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The energy of the photon is not enough to remove the
electron  nothing happens
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The energy of the photon is just enough to remove the
electron: the photon’s energy equals the ionization energy
 the electron leaves the atom without any Ek

The energy of the photon is larger than the ionization
energy  the electron leaves the atom with Ek
Exercise:
1.
2.
3.
4.
5.
What is the work function for the photosurface (in joules)?
What is the energy of the “green” photoelectron?
What is the speed of the “green” photoelectron?
What is the energy of the “blue” photoelectron?
What is the speed of the “blue” photoelectron?
Exercise:
1.
2.
3.
4.
5.
What is the work function for the photosurface (in joules)?
What is the energy of the “green” photoelectron?
What is the speed of the “green” photoelectron?
What is the energy of the “blue” photoelectron?
What is the speed of the “blue” photoelectron?
Light: wave or particle
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The photon has an energy given by E = hf
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But if it is considered a particle it also carries momentum
p=mv
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According to Einstein
E = m c2 ↔ m = E /c2
So,
p = (E /c2) c ↔ p = E / c
p = hf /c
p = h /λ
Light: wave or particle
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Light can behave as a particle and the photoelectric effect is
evidence for that fact.
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But if we do Young’s double slit experiment so that make
photons of light go through the slits one at each time, the
photon will produce an interference pattern.
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Somehow, even when light behaves like a particle it
conserves its wave properties.
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So, we talk about wave-particle duality.
De Bröglie’s wavelength
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In 1923, Louis de Bröglie suggested that if
light can behave as a particle then particles
could have a wave associated to them.
Louis de Broglie
The
wave-particle duality or duality of matter can be applied
to matter and energy.
All
particles have a wave associated to them so that
λ=h/p
Big
particles have a wavelength so small that it can’t be
measured.
But
small particles, like electrons, would have a wavelength
that is possible to be measured.
The electron as a wave
To
prove that the electron behaves like a wave it must have wave properties,
such diffraction. To make an electron diffract around an obstacle of size d, its
wavelength λ must be comparable to or bigger that d.
electron of mass 9.1x10-31kg and speed of 105 m/s will have a wavelength
λ = 7.2x10-9m.
An
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To have an obstacle with this size we must look at the structure of
crystals. The typical distance between atoms in a crystal is of the order
of 10-8m. When electrons are made to pass through crystals, they do
diffract thus proving its wave nature.
Davisson and Germer experiment
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In this experiment, electrons
of kinetic energy 54eV were
directed at a surface ok
nickel where a single crystal
had been grown and were
scattered by it.
Using the Bragg formula and
the known separation of the
crystal atoms allowed the
determination of the
wavelength which has then
seen to agree with the De
Broglie formula.
Structural analysis by electron
diffraction