If Planck’s constant is denoted by h and electronic charge by e, then photoelectric effect allows determination of:

In photoelectric effect we can know only the stopping potential.
Stopping potential $=(\frac{hc}{\lambda }-\varphi )/e$
                                $=\frac{hc}{e\lambda }-\varphi /e$
So, we can determine only $h/e$ because we know stopping potential, frequency and work function of the metal.