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.
How can we explain these observations? Look at the diagram
Question 1 Which one of the photoelectrons has the most kinetic energy? Why?
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.
Question 2 Which one of the photoelectrons has the least kinetic energy? Why?
We can summarise these findings in three rules, the laws of photoelectric emission.
 The number of electrons emitted per second depends on the intensity of the radiation.
 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.
 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.
Question 3 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 }JThis gives us a photon energy of 8.96 × 10^{19 }J
8.96 × 10^{19 }J = 6.6 × 10^{34} Js × f f = 8.96 × 10^{19 }J ÷ 6.6 × 10^{34} Js = 1.36 × 10^{15} Hz

What wavelength is this? 
Use the wave equation c = fll = 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.
Question 4 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.
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 + 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 – F
So the gradient is Planck’s constant, h.
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 =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} Js = 7.25 ´ 10^{14} Hz Use the wave equation to work out the wavelength: l = c/f = 3 ´ 10^{8} m/s ¸ 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 ´ 10^{8} m/s ¸ 350 ´ 10^{9} m = 8.57 ´ 10^{14} Hz Now put this into the photoelectric equation: E_{max} = hf – F = (6.63 ´ 10^{34} Js ´ 8.57 ´ 10^{14} 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 v^{2} = 2E_{k}/m = 2 ´ 8.8 ´ 10^{20} J ¸9.11 ´ 10^{31} kg = 1.93 ´ 10^{11} m^{2}/s^{2} Þ v = Ö(1.93 ´ 10^{11} m^{2}/s^{2}) = 4.4 ´ 10^{5} m/s Even these low energy electrons move like greased lightning. 
Question 5
(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 (E_{1}) and makes a transition to a lower level (E_{2}), then the energy of the photon given out can be worked out with the equation:
E = E_{1} – E_{2}
Since E = hf, we can rewrite this as:
hf = E_{1} – E_{2}
Let’s have a look at a worked example:
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 ´ 10^{8} m/s = 1.03 ´ 10^{7 }m = 107 nm
E 1.93 ´ 10^{18} J 
Question 6 (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.
He combined the following equations:
 Energy of photons: E = hf
 Einstein’s mass equivalence: E = mc^{2}
Therefore hf = mc^{2}.
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.
Question 7 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.