QuestionUltraviolet light of wavelength 250 nm is shone onto the metal cathode of a photocell. The stopping potential is found to be 1.80 V.(a) Explain what is meant by the stopping potential. (2)
(b) Calculate the maximum kinetic energy of the photoelectrons. (2)
(c) Calculate the work function of the metal, in J and in eV. (3)Answer(a) The minimum potential difference, making the anode negative relative to the cathode, (1) that stops even the most energetic photoelectrons from reaching the anode, so the current falls to zero. (1)(b) The work done against the field by the fastest electrons equals their maximum kinetic energy:Ek(max) = eVs = 1.60 × 10-19 × 1.80 = 2.88 × 10-19 J
(2)(c) f = c / λ = 3.00 × 108 ÷ 250 × 10-9 = 1.20 × 1015 Hzhf = 6.63 × 10-34 × 1.20 × 1015 = 7.96 × 10-19 J (1)φ = hf − Ek(max) = 7.96 × 10-19 − 2.88 × 10-19 = 5.08 × 10-19 J
(1)In electronvolts, using the unrounded value 5.076 × 10-19 J: φ = 5.076 × 10-19 ÷ 1.60 × 10-19 = 3.17 eV (1)Examiner tips- The stopping potential measures the maximum kinetic energy.
- Increasing the intensity of the light does not change the stopping potential; it only increases the current.
(b) Calculate the maximum kinetic energy of the photoelectrons. (2)
(c) Calculate the work function of the metal, in J and in eV. (3)Answer(a) The minimum potential difference, making the anode negative relative to the cathode, (1) that stops even the most energetic photoelectrons from reaching the anode, so the current falls to zero. (1)(b) The work done against the field by the fastest electrons equals their maximum kinetic energy:
(2)(c) f = c / λ = 3.00 × 108 ÷ 250 × 10-9 = 1.20 × 1015 Hzhf = 6.63 × 10-34 × 1.20 × 1015 = 7.96 × 10-19 J (1)
(1)In electronvolts, using the unrounded value 5.076 × 10-19 J: φ = 5.076 × 10-19 ÷ 1.60 × 10-19 = 3.17 eV (1)Examiner tips- The stopping potential measures the maximum kinetic energy.
- Increasing the intensity of the light does not change the stopping potential; it only increases the current.