Questions on the Photoelectric Effect

Q3.

a)

(i) Explain the meaning of the term 'work function' of a metal.

The work function of a metal is the minimum energy required to eject a (photo)electron from the metal surface.

In this mark scheme the examiner did nnot require the term 'surface' for a mark - but it is a good idea to always include it!

(ii) State what you would need to change in an experiment to investigate the effect of the work function on the photoelectric effect.

You would need to change the metal that the cathode was made of.

(3 marks)

(b) Experiments based on the photoelectric effect support the particle theory of light. State one conclusion drawn from these experiments and explain how it supports the particle theory.

Conclusion 1

Light of a frequency below a particular value, called the threshold frequency, does not release electrons. This happens because photons carry quanta of energy and electrons either absorb or reject a whole photon. Multiple photon are not absorbed by the electron in order to gain enough energy to escape from the metal, therefore lower energy photons would not be able to energise the electron so it could escape - a single photon had to do it.

Conclusion 2

Electrons are emitted immediately the light hits the metal surface .This happens because photons carry quanta of energy and electrons either absorb or reject a whole photon. Multiple photon are not absorbed by the electron in order to gain enough energy to escape from the metal, therefore if the photons are of high enough energy for emission to occur it will happen straight away - the process cannot occur due to a gradual build-up of energy..

(2 marks)

(c) Monochromatic light of wavelength 4.80 × 10–7 m falls onto a metal surface which has a work function of 1.20 × 10–19 J. Calculate

(i) the energy, in J, of a single photon of this light,

E = hf = hc/λ

E = (6.63 x 10-34 x3.0 x 108)/4.80 × 10–7

E = 4.144 x 10-19

E = 4.14 x 10-19 J

 

(ii) the maximum kinetic energy, in J, of an electron emitted from the surface.

hf = Φ + EkMAX

EkMAX = hf - Φ

EkMAX = 4.14 x 10-19 - 1.20 × 10–19

EkMAX = 2.94 × 10–19J

(5 marks)

(Total 10 marks)