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Date May 2018 Marks available 2 Reference code 18M.2.HL.TZ1.8
Level Higher level Paper Paper 2 Time zone 1
Command term Suggest 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.

M18/4/PHYSI/HP2/ENG/TZ1/08

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.

[2]
a.

Calculate, in J, the maximum kinetic energy of the emitted electrons.

[2]
b.i.

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.

[2]
b.ii.

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.

[1]
b.iii.

On the diagram, draw and label the equipotential lines at –0.4 V and –0.8 V.

[2]
c.i.

An electron is emitted from the photoelectric surface with kinetic energy 2.1 eV. Calculate the speed of the electron at the collecting plate.

[2]
c.ii.

Markscheme

E1 = –13.6 «eV» E2 = –  13.6 4 = –3.4 «eV»

energy of photon is difference E2E1 = 10.2 «≈ 10 eV»

 

Must see at least 10.2 eV.

[2 marks]

a.

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».

b.i.

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]

b.ii.

4.9 «V»

 

Allow 5.1 if 10.2 is used in (b)(i).

Ignore sign on answer.

[1 mark]

b.iii.

two equally spaced vertical lines (judge by eye) at approximately 1/3 and 2/3

labelled correctly

M18/4/PHYSI/HP2/ENG/TZ1/08.c.i/M

[2 marks]

c.i.

kinetic energy at collecting plate = 0.9 «eV»

speed = « 2 × 0.9 × 1.6 × 10 19 9.11 × 10 31 » = 5.6 × 105 «ms–1»

 

Allow ECF from MP1

[2 marks]

c.ii.

Examiners report

[N/A]
a.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
b.iii.
[N/A]
c.i.
[N/A]
c.ii.

Syllabus sections

Additional higher level (AHL) » Topic 12: Quantum and nuclear physics » 12.1 – The interaction of matter with radiation
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Additional higher level (AHL) » Topic 12: Quantum and nuclear physics
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