DP Physics (first assessment 2025)

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Question 21M.2.HL.TZ2.9

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Date May 2021 Marks available [Maximum mark: 6] Reference code 21M.2.HL.TZ2.9
Level HL Paper 2 Time zone TZ2
Command term Calculate, Explain, Outline Question number 9 Adapted from N/A
9.
[Maximum mark: 6]
21M.2.HL.TZ2.9

In an experiment to demonstrate the photoelectric effect, monochromatic electromagnetic radiation from source A is incident on the surfaces of metal P and metal Q. Observations of the emission of electrons from P and Q are made.

The experiment is then repeated with two other sources of electromagnetic radiation: B and C. The table gives the results for the experiment and the wavelengths of the radiation sources.

(a.i)

Outline the cause of the electron emission for radiation A.

[1]

Markscheme

photon transfers «all» energy to electron 

(a.ii)

Outline why electrons are never emitted for radiation C.

[1]

Markscheme

photon energy is less than both work functions
OR
photon energy is insufficient «to remove an electron»


Answer must be in terms of photon energy.

(a.iii)

Outline why radiation B gives different results.

[1]

Markscheme

Identifies P work function lower than Q work function

(b)

Explain why there is no effect on the table of results when the intensity of source B is doubled.

[1]

Markscheme

changing/doubling intensity «changes/doubles number of photons arriving but» does not change energy of photon

(c)

Photons with energy 1.1 × 10−18 J are incident on a third metal surface. The maximum energy of electrons emitted from the surface of the metal is 5.1 × 10−19 J.

Calculate, in eV, the work function of the metal.

[2]

Markscheme

5.1×10-19=1.1×10-18-ϕ ✓

work function = «(11.0-5.1)×10-191.6×10-19= » 3.7 «eV»  ✓


Award [2] marks for a bald correct answer.