22 17 In 1905 Einstein published his equation for the photoelectric effect. In 1916 Millikan demonstrated that the maximum kinetic energy of photoelectrons is consistent with Einstein’s equation. *(a) Discuss the extent to which our current understanding of observations of the photoelectric effect supports the idea that light behaves as photons rather than as waves. 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Turn over 23 (b) Millikan used his data to obtain a value of the Planck constant. The following graph of maximum kinetic energy of photoelectrons against frequency was produced from his data for the photoelectric effect using lithium. 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0 11 10 9 8 7 6 5 Frequency / 1014 Hz Maximum kinetic energy / 10−19 J Millikan suggested that the uncertainty from his results for lithium was as little as 1%. Determine whether the value of the Planck constant obtained from this graph is within 1% of the value stated on the data sheet for this examination paper. (3) .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. 24 (c) Millikan’s experiments involved using different frequencies of light. These were obtained using a mercury vapour lamp which produced an emission spectrum with a specific number of known frequencies. The diagram shows some energy levels for a mercury atom. 0 eV −1.56 eV −1.57 eV −2.48 eV −3.71 eV −4.95 eV −5.52 eV −5.74 eV −10.38 eV Not to scale Determine which transition from the −3.71 eV energy level would produce light of wavelength 6.1 × 10−7 m. 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Transition from −3.71 eV to .................................................................... Turn over 25 (d) Millikan used a device known as a monochromator to ensure that a single wavelength of light was used to illuminate the surface of the lithium. A monochromator separates wavelengths using a diffraction grating. Calculate the angle at which a diffraction grating would produce the most intense line at a single wavelength of 6.1 × 10−7 m. number of lines per mm for grating = 600 mm−1 (3) ................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................... Angle = .................................................................... (Total for Question 17 = 16 marks)