The energy of the incident photon is given by:\[\frac{hc}{\lambda} = eV_s + \phi\]Substitute the given values:\[\frac{1240}{300} \, \text{eV} - 2.13 \, \text{eV} = eV_s\]\[4.13 \, \text{eV} - 2.13 \, \text{eV} = eV_s\]\[V_s = 2 \, \text{V}\]