Quantum Tutorial 2  The Photoelectric Effect
Contents 
Photons and Quantum Physics
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 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.
How can we explain these observations? Look at the diagram:
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? 
Photoelectric Emission
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. A voltage of 5 volts gives rise to an electron energy of 5 eV. 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
of 5 V, the work is done is 5 eV or 8.0 × 10^{19 }J.
If the stopping voltage for a photoelectron is 3 V, what is the kinetic energy in eV and joules? 
Threshold Frequency
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; weak gamma rays will.
Worked Example
What
is the threshold frequency of a metal whose photoelectrons are stopped by
a stopping voltage of 5.6 V? 
Answer
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

Worked Example What wavelength is this? 
Answer Use the wave equation c = fl
l = 3 × 10^{8} m/s ÷ 1.36 × 10^{15} 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.
Confusion here is a very common bear trap. 
We can use a model to explain this. Consider a rugby player about to take a place kick. (I was hopeless at rugby. I learned that fairies existed  "Get in there ya fairy!")
Image from WikiHow
If the kicker kicks the ball with too little energy, it will stay where it is.
If he kicks the ball with just enough energy, the ball will roll off the stand.
If
there is more than enough energy, the ball will fly off (over the bar?).
Einstein’s Photoelectric Equation
Summing up what we have learned already:
When photoelectrons are removed from a metal surface, a certain amount of work has to be done in removing them.
The photoelectrons will lose some of their kinetic energy in order to escape the attractive field of the positively charged ions.
The work required to remove the photoelectron is called the work function.
It is given the physics code
f (phi
 a Greek letter ‘f’) and is measured in joules, or electron volts.
Sometimes, depending the syllabus or text book, the code is written as a
capital Phi, F.
We find that the gradient of this graph is constant, regardless of the metal.
It is tempting to relate the work function to the molar ionisation energy. If you convert the work function to kilojoules per mole, you will find that it is somewhat lower than the ionisation energy per mole. This is because the ionisation energy in chemistry is worked out using metals that have been turned into gases. The photoelectric effect involves solid metals which have free electrons which are more easily removed. 
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.
This is illustrated in the diagram below:
Energy
of Photon = work done to remove electron + kinetic energy of the electron
In
code:
E =
f + E_{k}
E =
f
+ 1/2 mv^{2}
We must note the following:
E_{k}
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:
E_{k} =
hf 
The work function can be worked out from the threshold frequency, f_{0}. At the threshold frequency, E_{k} = 0. Therefore:
0 = hf_{0}  f
f = hf_{0}
Therefore we can rearrange:
Where:
f_{0} – threshold frequency (Hz)
f – work function (J)
h – Planck’s Constant (J s)
The Photoelectric Effect Equation is usually written like this:
Worked 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. 
Answer
(a)
Work out the work function in joules: f = 3.0 eV × 1.6 × 10^{19} J eV^{1} = 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 = hf_{0} where f_{0} is the threshold frequency. Since the energy given by the photon is the work function f, we can rewrite the equation as f = hf_{0}.
Rearranging: f_{0} = f/h = 4.8 × 10^{19} J ÷ 6.63 × 10^{34} J s = 7.24 × 10^{14} Hz
Use the wave equation to work out the wavelength:
l =
c/f = 3.0 ×
10^{8} m s^{1} ¸
7.25 ×
10^{14} Hz = 4.14 ×
10^{7} m = 414 nm 
(b)
Use
E_{max} = hf 
f
First work out the frequency of the 350 nm light:
f = c/l = 3.0
×
10^{8} m s^{1}
¸
350 ×
10^{9} m = 8.57 ×
10^{14} Hz
Now put this into the photoelectric equation:
E_{max} = hf  f
= (6.63 ×
10^{34} J s ×
8.57 ×
10^{14} Hz)  4.8 ×
10^{19} J = 8.82
×
10^{20} J
This is equivalent to 0.55 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
v^{2} = 2E_{k}/m
= 2 ×
8.82 ×
10^{20} J ¸ 9.11
×
10^{31} kg = 1.94 ×
10^{11} m^{2} s^{2} Þ v = Ö(1.93 × 10^{11} m^{2} s^{2}) = 4.40 × 10^{5} m s^{1} (= 4.4 × 10^{5} m s^{1} to 2 significant figures)
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? 