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Date May 2017 Marks available 2 Reference code 17M.2.HL.TZ1.9
Level Higher level Paper Paper 2 Time zone Time zone 1
Command term Determine Question number 9 Adapted from N/A

Question

Two observations about the photoelectric effect are

Observation 1: For light below the threshold frequency no electrons are emitted from the metal surface.

Observation 2: For light above the threshold frequency, the emission of electrons is almost instantaneous.

The graph shows how the maximum kinetic energy Emax of electrons emitted from a surface of barium metal varies with the frequency f of the incident radiation.

Explain how each observation provides support for the particle theory but not the wave theory of light.

[4]
a.

Determine a value for Planck’s constant.

[2]
b.i.

State what is meant by the work function of a metal.

[1]
b.ii.

Calculate the work function of barium in eV.

[2]
b.iii.

The experiment is repeated with a metal surface of cadmium, which has a greater work function. Draw a second line on the graph to represent the results of this experiment.

[2]
c.

Markscheme

Observation 1:
particle – photon energy is below the work function
OR
E = hf and energy is too small «to emit electrons»
wave – the energy of an em wave is independent of frequency


Observation 2:
particle – a single electron absorbs the energy of a single photon «in an almost instantaneous interaction»
wave – it would take time for the energy to build up to eject the electron

a.

attempt to calculate gradient of graph = «\(\frac{{4.2 \times {{10}^{ - 19}}}}{{6.2 \times {{10}^{14}}}}\)»

\( = 6.8 - 6.9 \times {10^{ - 34}}\) «Js»

 

Do not allow a bald answer of 6.63 x 10-34 Js or 6.6 x 10-34 Js.

b.i.

ALTERNATIVE 1
minimum energy required to remove an electron «from the metal surface»

ALTERNATIVE 2
energy required to remove the least tightly bound electron «from the metal surface»

b.ii.

ALTERNATIVE 1
reading of y intercept from graph in range 3.8 − 4.2 × 10–19 «J»
conversion to eV = 2.4 – 2.6 «eV»

ALTERNATIVE 2
reading of x intercept from graph «5.8 − 6.0 × 1014 Hz» and using hf0 to get 3.8 − 4.2 × 10–19 «J»
conversion to eV = 2.4 – 2.6 «eV»

b.iii.

line parallel to existing line
to the right of the existing line

c.

Examiners report

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

Syllabus sections

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