A-LevelPhysicsQuantum physics9702/43 · Oct/Nov 2019 · Q1112 Marks

(a) With reference to the photoelectric effect, state what is meant by work function energy. (b) The work function energy of a clean metal surface is 5.5 × 10⁻¹⁹ J. Electromagnetic radiation of wavelength 280 nm is incident on the metal surface. The metal is in a vacuum.

(a)With reference to the photoelectric effect, state what is meant by work function energy.[2]
(b)(i) 1.Calculate: the photon energy[2]
(b)(i) 2.Calculate: the maximum speed V_MAX of the electrons emitted from the surface.[3]
(b)(ii)Explain why most of the emitted electrons will have a speed lower than V_MAX.[1]
(c)The electromagnetic radiation incident on the metal surface may change in intensity or in frequency. Complete Fig. 11.1 by inserting either 'increases' or 'decreases' or 'no change' to describe the effects of the changes shown on the maximum speed and on the rate of emission of electrons. [Figure 11.1][4]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.

[M1] (a) energy (of photon) required to remove electron
[A1] (a) from a surface
[C1] (b)(i) 1. photon energy = hc / λ
[A1] (b)(i) 1. = 7.1 x 10⁻¹⁹ J
[C1] (b)(i) 2. electron energy = (7.1 – 5.5) × 10⁻¹⁹ J
[C1] (b)(i) 2. ½ x 9.11 x 10⁻³¹ x v² = (7.1 – 5.5) × 10⁻¹⁹
[A1] (b)(i) 2. v = 5.9 x 10⁵ m s⁻¹
[B1] (b)(ii) energy is required to bring electron to the surface
[B1] (c) reduced intensity at constant frequency: maximum speed = no change
[B1] (c) reduced intensity at constant frequency: rate of emission = decreases
[B1] (c) increased frequency at constant intensity: maximum speed = increases
[B1] (c) increased frequency at constant intensity: rate of emission = no change

On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.