Question

The graphs below show the variation of stopping potential versus frequency of incident radiation for metals A and B. 

UV radiation of appropriate wavelength is allowed to fall on both the metals. Which metal will emit photoelectrons with higher maximum kinetic energy (E_max)? Give a reason.

Answer 1

1. From the graph, the threshold frequency of Metal A is less than that of Metal B (f_0A ​< f_0B​).
2. Since the work function is given by Φ = hf₀, Metal A has a smaller work function than Metal B.
3. According to Einstein’s photoelectric equation, E_max = hf − Φ.
4. The incident UV radiation is the same for both metals, so hf is constant.
5. As Metal A has a smaller work function, more energy remains as kinetic energy of emitted electrons.

Therefore, Metal A emits photoelectrons with a higher maximum kinetic energy.