Topic 4 Waves in Boxes
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Key Words Photoelectron, Kinetic Energy, Work Function, Energy Levels, de Broglie Wavelength |
Albert
Einstein developed the theory further to study how atoms interacted with
photons. He produced the notion of quantum
physics, in which electromagnetic radiation has a particulate nature. The essential points of quantum theory are:
All
electromagnetic radiation is emitted in tiny bursts of energy called photons
Photons
travel in one direction only and in a straight line
When an
atom emits a photon its energy changes by the energy of the photon.
Energy
contained in a photon is given by E = hf.
Detailed study called for more sophisticated experiments that used apparatus like this

Details of this experiment are NOT needed for the AQA Module 1 exam. However to understand the results, we need to be aware of what goes on in the experiment:
The photocathode is given a positive voltage, and the photoanode a negative voltage.
This means that photoelectrons (electrons released by interaction with a photon. One photon releases one electron) are repelled from the anode.
If the electrons have lots of kinetic energy, they can overcome the repulsive force.
We turn up the reverse voltage until the electrons with the most kinetic energy are just repelled. The voltage is called the stopping voltage. We can see what is happening in this diagram

The totally unexpected result is that the
maximum
kinetic energy of the photoelectrons is exactly the same regardless of the
intensity of the illumination. However
dim or bright the light, the maximum kinetic energy is the same.

| Which one of the photoelectrons has the most kinetic energy? Why? |
It is important to understand that each photon can eject only one photoelectron. So if we can count the number of photoelectrons, we know the number of photons.
Although
the diagram is a simplification as to what really happens, we can see that the
photoelectrons are released with a range
of kinetic energies. The
lowest kinetic energy is where the electron just manages to crawl out.
It will be hauled back pretty quickly by the electrostatic forces.
| Which one of the photoelectrons has the least kinetic energy? Why? |
We
can summarise these findings in three rules, the laws of photoelectric emission.
1.
The
number of electrons emitted per second depends on the intensity of the
radiation.
2.
The
photoelectrons have a range of energy, from zero to a maximum value.
The maximum value is determined by the frequency of the radiation, not
the intensity.
3.
A minimum
value for the frequency is needed, the threshold frequency.
The
maximum kinetic energy has the same value
in eV as the stopping voltage. This
stands to reason. We know that
energy = charge × voltage, and that the electron carries a single electronic
charge (1e = 1.6 × 10-19 C). So if that charge moves through a potential difference, that
amount of work is done.
| If the stopping voltage for a photoelectron is 3 V, what is the kinetic energy in eV and joules? |
The graph shows how the energy of the photoelectrons depends on the frequency (colour) of the light:

There
has to be a
threshold frequency below
which no photoelectrons are emitted, regardless of brightness.
Therefore radio waves, however strong, will NEVER affect photographic
film; gamma rays will.
|
What
is the threshold frequency of a metal whose photoelectrons are stopped by
a stopping voltage of 5.6 V? |
| The
maximum kinetic energy is 5.6 eV = 5.6
× 1.6 × 10-19 = 8.96 × 10-19 J This
gives us a photon energy of 8.96 × 10-19 J
|
| What wavelength is this? |
| Use the wave equation c =
fl
l = 3 × 108 m/s ÷ 1.36 × 1015 Hz = 2.20 × 10-7 m = 220 nm |
Physicists tend to use nanometres to measure wavelength of light, so red light has a wavelength of 600 nm. 600 nm = 600 × 10-9 = 6 × 10-7 m.
Failure
to convert correctly from nanometres to metres is a common bear trap.

| A metal gives out photoelectrons that have a stopping voltage of 2.6 V. Will light of wavelength 615 nm cause photoelectrons to be ejected? |
When you answer
this kind of question, you need to be very careful about what you say when
discussing wavelengths. A photon
with a long wavelength carries less energy than one with a short wavelength.
So if the wavelength of a photon is longer than the wavelength suggested
by the threshold frequency, photoelectrons will not be ejected.

Einstein’s Photoelectric Equation

When
photoelectrons are removed from a metal surface, a certain amount of work has to
be done in removing them. Therefore
the photoelectrons will lose some of their kinetic energy in order to escape the
attractive field of the positively charged nuclei.
The work required to remove the photoelectron is called the work
function. It is given the
physics code F
(Phi - a Greek capital letter ‘Ph’) and is measured in joules, or electron
volts.
The
energy received from a photon is split into:
The work
necessary to separate the electron from the metal (the work function)
The
kinetic energy.
Energy
of Photon = work done to remove electron + kinetic energy of the electron
In
code:
E = F + Ek
E =
F
+ 1/2 mv2
We must note the following:
Ek
is the maximum kinetic energy (the
charge × stopping voltage), i.e. the kinetic energy of the
fastest electrons. We are not
interested in slower electrons.
The
maximum kinetic energy is dependent only on the frequency, NOT the intensity.
A more intense beam produces more photons per second, but each photon has
the same energy.
We can work out the work function of any metal by plotting the maximum energy against the frequency

We find that the gradient of this graph is constant, regardless of the metal. The equation of the graph is:
Ek =
hf - F
|
Example A metal surface has a work function of 3.0 eV and is illuminated with radiation of wavelength 350 nm. Work out: (a) The maximum wavelength that causes photoelectric emission (b) The maximum kinetic energy of the photoelectrons (c) The speed of the photoelectrons. |
|
(a)
Work out the work function in joules: F = 3.0 eV ´ 1.6 ´ 10-19 J/eV = 4.8 ´ 10-19 J. Now work out the frequency that this corresponds to. The minimum frequency is the frequency at which a photon will just release an electron. So we use the equation E =hf0 where f0 is the threshold frequency. Since the energy given by the photon is the work function F, we can rewrite the equation as F = hf0. Rearranging: f0 = F/h = 4.8 ´ 10-19 J ¸ 6.63 ´ 10-34 Js = 7.25 ´ 1014 Hz Use the wave equation to work out the wavelength: l
= c/f = 3 ´
108 m/s ¸
7.25 ´
1014 Hz = 4.14 ´
10-7 m = 414 nm |
|
(b)
Use Emax = hf -
F First work out the frequency of the 350 nm light: f = c/l = 3
´
108 m/s ¸
350 ´
10-9 m = 8.57 ´
1014 Hz Now put this into the photoelectric equation: Emax = hf - F
= (6.63 ´
10-34 Js ´
8.57 ´
1014 Hz) - 4.8 ´
10-19 J = 8.8 ´
10-20 J This is equivalent to 0.54 eV. It is perfectly acceptable to express your answers in eV or joules.
|
|
(c)
Now we use the kinetic energy of the electron to find out its
speed: Mass of an electron = 9.11 ´ 10-31 kg v2 = 2Ek/m
= 2 ´
8.8 ´
10-20 J ¸9.11
´
10-31 kg = 1.93 ´
1011 m2/s2 Þ v = Ö(1.93 ´ 1011 m2/s2) = 4.4 ´ 105 m/s Even these low energy electrons move like greased lightning. |

|
(a) The Work Function of potassium is 3.52 × 10-19 J. What is meant by this statement? (b) When radiation of a suitable frequency falls on a potassium surface, photoelectrons are emitted. What is the minimum frequency at which this can occur? (c) What is the maximum speed of the photoelectrons emitted when radiation of 400 nm falls on the potassium surface? |
Energy Levels
in Atoms
Atoms can interact with photons of lower energy than is required to remove electrons from them. The photons we looked at in the photoelectric effect could remove the electrons from very reactive metals like caesium. Photons can interact with other atoms to give them extra energy, which makes them excited.
When we heat a gas or pass an electric current through it we can make it glow. We have ionised the gas. If we look at the glowing gas through a spectrometer, we see the spectrum of the gas which is distinctive for that gas. Unlike the spectrum of the Sun, in which we see all the colours of the rainbow, we only see certain colours, while others are absent. We call this kind of spectrum a line emission spectrum. The colours are discrete wavelengths

When a gas is ionised, one or more outer electrons are ripped off. The molecule has become positive. It will recombine with an electron and lose energy, giving that energy in the form of a photon. Other atoms may not have been ionised, but are still in a very excited state. The atoms have interacted with the photon and the electrons have moved to a higher energy level.
About a microsecond later, the electrons lose their energy as a photon and return to the stable state, called the ground state. The important thing to remember is that electrons can only exist at permitted energy levels. It’s like a person standing on a ladder; he can exist at one rung up, two rungs, etc., but NOT at a height of 1.5 rungs.
As we consider energy levels in atoms, we will look at hydrogen which fits this model well. (Hydrogen has one electron.) More complex atoms with several electrons do not.
If we look at a spectrum of hydrogen, we find lines at several discrete wavelengths.
Each line represents the energy of a photon as the electron makes a transition from a higher energy level to a lower. This we can show in a diagram below:
The electron does a job of work in releasing a photon; it has lost potential energy. Therefore we start at the highest level which we give a value of zero. Therefore the electron falls from the zero point to the –3.41 eV level. The more negative the level, the lower the energy level.
The highest energy level is where ionisation occurs. The lowest level is the ground state.
Electrons can make transitions from any energy level to any other:

These transitions give us photons in the visible spectrum. In fact, the ground state is at –13.6 eV. This is the ionisation energy of hydrogen, the energy required to strip an electron from the atom.

We need to be aware of the following points:
The
lowest level (-13.6 eV) is the ground state.
This is the normal configuration of the atom. Energy must be put in to raise the electron to other levels.
The
highest level is the ionisation energy.
Energy
levels are not evenly spaced.
We can quantify this in an equation. If
an electron is at an excited level (E1)
and makes a transition to a lower level (E2),
then the energy of the photon given out can be worked out with the equation:
E = E1 – E2
Since
E = hf, we can rewrite this as:
hf = E1 – E2
|
What is the wavelength of photons of light given out by the transition from –1.51 eV to the ground state (-13.6 eV)? |
| Energy given out = -1.51 eV – (-13.6 eV) = 12.09 eV |
| Energy in joules = 12.09 eV ´ 1.6 ´ 10-19 J/eV = 1.93 ´ 10-18 J |
|
Use
l = hc
= 6.63 ´
10-34 Js ´
3.0 ´
108 m/s
= 1.03 ´ 10-7 m = 107 nm E 1.93 ´ 10-18 J |
|
(a) An excited atom loses its energy quickly. How does it do this? (b) What is the frequency of a photon given out by a transition from -0.85 eV to -1.51 eV? |
Wave Behaviour of Particles
The
Belgian physicist de Broglie (pronounced ‘de Broy’) reasoned that if waves have a
particulate properties, it was reasonable to suppose that particles had wave
properties. He devised the
relationship, which states that particles have wave properties.
It is the logical extension of the particulate nature of electromagnetic
wave phenomena.
Energy of
photons:
E = hf
Einstein’s
mass equivalence:
E
= mc2
Therefore
hf = mc2.
Now
f = c/l
So mc = h/l
The
term mc is mass ´ velocity, which is momentum.
We give momentum the code p.
We can rewrite the equation as
l = h/p
or
l
= h/mv
Therefore
every particle with a momentum has an associated de Broglie wavelength, even
something as absurd as a car travelling at 20 m/s.
| What is the de Broglie wavelength of an electron travelling at 2 × 10 6
m/s? |
Electrons can be shown to have wave properties by the simple use of an electron diffraction tube. A slice of carbon is placed in a beam of electrons so that the electrons diffract.

We
need to note a couple of points:
l
is the de Broglie wavelength
Strictly
speaking we should count the mass and speed as relativistic.
As the speed of particles approaches the speed of light, the mass
increases as kinetic energy is turned into mass.
We will not worry about this at this stage.
The
wave properties of electrons have led to the development of the electron
microscope, which allows magnifications much bigger than was ever possible
with the light microscope. A good
light microscope can magnify up to 1000 times.
The electron microscope can magnify up to about 1 million times, and can
reveal the existence of individual atoms. The
electron beams are focused by magnets just like the lenses on a microscope.
|
Suggested Websites |
| http://lectureonline.cl.msu.edu/~mmp/kap28/PhotoEffect/photo.htm |
| http://www.colorado.edu/physics/2000/quantumzone/photoelectric2.html |

| Presentation | Quantum Theory | ||
|
Now try Topic Quiz 4 |
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