Calculating the proportion of incident energy transferred to kinetic energy in the photoelectric effect

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?

• When professors ask students about the "maximum proportion of the incident energy that could be transformed into the kinetic energy of the photoelectrons", they are vastly more likely to be asking about $\frac{K_2}{E_2}$ and that there is never both $E_1$ and $E_2$ being shone on the metal at the same time. Your question is just very weird. Commented Feb 26 at 19:34
• hyperphysics.phy-astr.gsu.edu/hbase/mod2.html Commented Feb 26 at 19:46
• @naturallyInconsistent Although it seems weird, this question definitely wants me to consider $E_1$ and $E_2$ both being shone on the metal at the same time. Commented Feb 26 at 20:32