Physics · Quantum physics · 7 marks
Light of wavelength 450 nm is shone onto a metal surface of work function 1.9 eV. (a) Show that photoelectrons are emitted. [3] (b) Calculate the maximum kinetic energy of an emitted electron, in joules. [2] (c) State and explain the effect on the photoelectric current of doubling the intensity. [2] h = 6.63 × 10⁻³⁴ J s · c = 3.00 × 10⁸ m s⁻¹ · e = 1.60 × 10⁻¹⁹ C
Mark scheme
- (a) E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(450 × 10⁻⁹)
- (a) E = 4.42 × 10⁻¹⁹ J = 2.76 eV
- (a) 2.76 eV > 1.9 eV, so the photon energy exceeds the work function and emission occurs
- (b) Eₖ(max) = hf − φ = 4.42 × 10⁻¹⁹ − (1.9 × 1.60 × 10⁻¹⁹)
- (b) Eₖ(max) = 1.38 × 10⁻¹⁹ J
- (c) The current doubles, because twice as many photons arrive per second so twice as many electrons are emitted
- (c) The maximum kinetic energy is unchanged, because it depends on frequency not intensity
Part (c) is the one that separates grades: intensity changes how many, frequency changes how fast. Saying both change is an automatic loss.