Light of two wavelengths, $λ_1$ and $λ_2$, is incident on a metal surface of work function $ϕ$. If there is an equal number of photons of each wavelength incident per second, what is the maximum proportion of the incident energy that could be transformed into the kinetic energy of the photoelectrons? I'm deliberately refraining from adding specific numbers to this problem as I want to understand the principles rather than specific computations.
My first thought is to calculate the energies of the two photons corresponding to the two wavelengths: $$E_1=\frac{hc}{λ_1}$$ $$E_2=\frac{hc}{λ_2}$$
Let's just say that $E_1 < ϕ$ and that one photon of each wavelength is incident per second to make the problem a little easier to understand.
If the energy of the photon is less than the work function, then no photoelectron will be released, so I can forget about $E_1$. For $E_2$, the kinetic energy of the emitted photoelectron can be calculated using: $$K_2=E_2-ϕ$$Therefore, the maximum proportion of the incident energy that could be transformed into the kinetic energy of the photoelectrons would be: $$\frac{K_2}{E_1+E_2}$$Have I understood the photoelectric effect correctly?
