| Date | May 2018 | Marks available | 1 | Reference code | 18M.2.HL.TZ1.8 |
| Level | Higher level | Paper | Paper 2 | Time zone | Time zone 1 |
| Command term | Write down | Question number | 8 | Adapted from | N/A |
Question
Hydrogen atoms in an ultraviolet (UV) lamp make transitions from the first excited state to the ground state. Photons are emitted and are incident on a photoelectric surface as shown.
The photons cause the emission of electrons from the photoelectric surface. The work function of the photoelectric surface is 5.1 eV.
The electric potential of the photoelectric surface is 0 V. The variable voltage is adjusted so that the collecting plate is at –1.2 V.
Show that the energy of photons from the UV lamp is about 10 eV.
Calculate, in J, the maximum kinetic energy of the emitted electrons.
Suggest, with reference to conservation of energy, how the variable voltage source can be used to stop all emitted electrons from reaching the collecting plate.
The variable voltage can be adjusted so that no electrons reach the collecting plate. Write down the minimum value of the voltage for which no electrons reach the collecting plate.
On the diagram, draw and label the equipotential lines at –0.4 V and –0.8 V.
An electron is emitted from the photoelectric surface with kinetic energy 2.1 eV. Calculate the speed of the electron at the collecting plate.
Markscheme
E1 = –13.6 «eV» E2 = – \(\frac{{13.6}}{4}\) = –3.4 «eV»
energy of photon is difference E2 – E1 = 10.2 «≈ 10 eV»
Must see at least 10.2 eV.
[2 marks]
10 – 5.1 = 4.9 «eV»
4.9 × 1.6 × 10–19 = 7.8 × 10–19 «J»
Allow 5.1 if 10.2 is used to give 8.2×10−19 «J».
EPE produced by battery
exceeds maximum KE of electrons / electrons don’t have enough KE
For first mark, accept explanation in terms of electric potential energy difference of electrons between surface and plate.
[2 marks]
4.9 «V»
Allow 5.1 if 10.2 is used in (b)(i).
Ignore sign on answer.
[1 mark]
two equally spaced vertical lines (judge by eye) at approximately 1/3 and 2/3
labelled correctly
[2 marks]
kinetic energy at collecting plate = 0.9 «eV»
speed = «\(\sqrt {\frac{{2 \times 0.9 \times 1.6 \times {{10}^{ - 19}}}}{{9.11 \times {{10}^{ - 31}}}}} \)» = 5.6 × 105 «ms–1»
Allow ECF from MP1
[2 marks]
