This will assume that either a photon has been absorbed or released by the electron. It will still work either way.
#E = hf = (hc)/lamda#, where E is the energy of the photon, h is Planck's constant (#6.63x10^(-34)J s)# f is tje frequency of light, c is the speed of light (which I'll take as #3.00x10^8ms^-1#), and #lamda# is the wavelength of the light.
#E = ((6.63*10^(-34))(3.00*10^8))/(7.464*10^(-6)) = 2.664791*10^(-20)J#
If you want your answer in eV, then #abs(e) = 1.6*10^(-19)C#, and as #V = E/Q#, where V is the potential difference, E is energy, and Q is #abs(e)#, you get #(2.664791*10^(-20))/(1.6*10^(-19)) = 0.166549438eV = 0.17eV#
