Exam Question 6

QuestionA student measures the maximum kinetic energy E_k(max) of photoelectrons emitted from a metal for light of several different frequencies f, and plots a graph of E_k(max) against f. The result is a straight line that crosses the frequency axis at 5.0 × 10¹⁴ Hz.(a) Use the photoelectric equation to explain why the graph is a straight line. (2)
(b) State what the gradient of the line represents. (1)
(c) Calculate the work function of the metal. (2)
(d) The experiment is repeated with a metal of larger work function. Describe how the new line compares with the first. (2)Answer(a) Rearranging hf = φ + E_k(max) gives:E_k(max) = hf − φ
This has the form y = mx + c, with h and φ both constant, so E_k(max) against f is a straight line. (2)(b) The gradient is the Planck constant, h. (1)(c) The intercept on the frequency axis is the threshold frequency, f₀ = 5.0 × 10¹⁴ Hz, where E_k(max) = 0. (1)φ = hf₀ = 6.63 × 10^-34 × 5.0 × 10¹⁴ = 3.3 × 10^-19 J
(1)(d) The new line is parallel to the first, because the gradient is still h. (1) It is shifted to the right, crossing the frequency axis at a higher threshold frequency. (1)Examiner tips- If the line is extended back to f = 0, it meets the E_k(max) axis at −φ. This gives a second way to find the work function.
- The line does not pass through the origin.