The Photoelectric Effect

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:

Detailed study called for more sophisticated experiments that used apparatus like this

Details of this experiment are NOT needed for the AQA Unit 1 exam.  However to understand the results, we need to be aware of what goes on in the experiment:

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? 


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. 

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.

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.

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 J

This 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 × 1015 Hz


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.

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 letter ‘ph’) and is measured in joules, or electron volts.

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:

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:

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

 So the gradient is Planck’s constant, h.


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.


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.

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?