1. Quantum Practice Paper

2. 1. Use data from the Data Sheet in this question.
(a) (i) Define the electron volt.
(ii) Show that the speed of an electron accelerated through a potential difference of 6.0kV is 4.6 × 107 m s–1.
(b) State what is meant by the duality of the nature of electrons.

3. 1. (a) (i) electronvolt is the energy gained (1) by an electron moved through a p.d of 1 V (1)
(ii) 1/2mv2 = eV (1)
v = 4.6 ×107 ms–1
(b) electrons can behave like waves and like particles (1)

4. 3. (a) (i) Explain the meaning of the term work function of a metal
(ii) State what you would need to change in an experiment to investigate the effect of the work function on the photoelectric effect.(3)
(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.

5. 3.(a)(i) the minimum energy (1) energy required to eject a (photo)electron (from the metal surface) (1)
(ii) changing the metal/cathode (1)
(b) conclusion: light below a threshold frequency does not release electrons (1) explanation: photons carry quanta of energy (1)
[or conclusion: electrons are emitted immediately the light hits the metal surface (1) explanation: photons carry quanta of energy (1)]

6. 3c) Monochromatic light of wavelength 4
3c) 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,

7. (c) (i) (use of E = hf= gives) E = (1)
4.14 × 10–19 (J)
(ii) hf = f + Ek (1) Ek = 4.14 × 10–19 – 1.20 × 10–19 (1) = 2.94 × 10–19 (J) (1) (allow C.E for value of E from (i))

8. Explain what is meant by (i) excitation, (ii) ionisation.
4. The diagram shows some of the energy levels of the mercury atom.
(a) When electrons collide with mercury atoms, the atoms may be excited or may be ionised.
Explain what is meant by
(i) excitation,
(ii) ionisation.
(b) Determine the lowest frequency of emitted radiation with reference to the energy levels in the diagram.

9. 4. (a)(i) an electron moves up from one energy level to another (1)
(ii) an electron is removed from an atom (1)
(b) (use of hf = E2 – E1 gives) f = (2.56 – 1.92) × 10–19 (1) /6.63 × 10–34 = 9.65 × 1013 Hz