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Question 18M.2.HL.TZ1.8

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Date May 2018 Marks available [Maximum mark: 11] Reference code 18M.2.HL.TZ1.8
Level HL Paper 2 Time zone TZ1
Command term Calculate, Draw, Show that, Suggest, Write down Question number 8 Adapted from N/A
8.
[Maximum mark: 11]
18M.2.HL.TZ1.8

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

(a)

Show that the energy of photons from the UV lamp is about 10 eV.

[2]

Markscheme

E1 = –13.6 «eV» E2 = –  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]

The photons cause the emission of electrons from the photoelectric surface. The work function of the photoelectric surface is 5.1 eV.

(b.i)

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

[2]

Markscheme

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.ii)

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]

Markscheme

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.iii)

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]

Markscheme

4.9 «V»

 

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

Ignore sign on answer.

[1 mark]

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.

(c.i)

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

[2]

Markscheme

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.ii)

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]

Markscheme

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]

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