The photoelectric effect shows light arriving in packets. Each electron gets the energy of one photon, as in The Photoelectric Effect; what is left after the work function becomes its kinetic energy. Each problem here is one short scene and a chain of questions. Answer them in order: each later question uses what you found before, so a slip early on carries through. Keep full precision between steps and round only at the end.
Key idea
$$E = hf = \frac{hc}{\lambda}, \quad hc = 1240\ \text{eV·nm}; \qquad K_{\max} = E - \phi; \qquad eV_s = K_{\max}.$$ Brighter light means more photons, so more electrons, but not faster ones.
Worked example: Ultraviolet on potassium
Light of $300\ \text{nm}$ falls on potassium ($\phi = 2.3\ \text{eV}$). Find (1) the photon energy, (2) $K_{\max}$ and (3) the stopping voltage.
- (1) $E = 1240/300 = 4.13\ \text{eV}$.
- (2) $K_{\max} = 4.13 - 2.3 = 1.83\ \text{eV}$.
- (3) $V_s = 1.83\ \text{V}$.
Answer: $4.13\ \text{eV}$, $1.83\ \text{eV}$, $1.83\ \text{V}$
Common mistake
Expecting brighter light to give faster electrons. The energy each electron gets depends only on the photon, so on the wavelength. Check: below the threshold frequency, no brightness gets any electron out.