Ultraviolet radiation of wavelength 200 nm is incident on a clean zinc plate. The work function of zinc is 4.3 eV.
(a) Calculate the maximum kinetic energy of the photoelectrons, in J. [4]
(b) Visible light of wavelength 500 nm does not release electrons from zinc, however intense it is. Explain why. [2]
- (a) $E = \dfrac{hc}{\lambda} = \dfrac{6.63\times10^{-34}\times3.00\times10^{8}}{200\times10^{-9}}$(1)
- (a) $E = 9.95\times10^{-19}$ J allow 9.9 × 10-19 J(1)
- (a) $\phi = 4.3 \times 1.60\times10^{-19} = 6.88\times10^{-19}$ J (allow 6.9 × 10-19 J)(1)
- (a) $E_{k(max)} = 9.95\times10^{-19} - 6.88\times10^{-19} = 3.1 \times 10^{-19}$ J allow 3.0 × 10-19 J if intermediate values are rounded to 2 s.f.(1)
- (b) photon energy at 500 nm (2.5 eV / $4.0\times10^{-19}$ J) is less than the work function(1)
- (b) one electron absorbs one photon, so increasing intensity only increases the number of photons, not the energy of each(1)
- Answer: (a) $3.1\times10^{-19}$ J (b) each photon has energy below the work function; one photon–one electron
- Examiner tip: Convert the work function to joules before subtracting — mixing eV and J is the classic error.