Maximum kinetic energies of photoelectrons emitted from a metal surface when illuminated by radiations of wavelengths `lamda_(1) and lamda_(2), " are " K_(1) and K_(2)`, respectively. If `lamda_(1) = 3 lamda_(2)`, show that, `K_(1) lt (K_(2))/(3)`

Explanation

If the work function of the metal is W, from Einstein's equation,
`K_(1) = (hc)/(lamda_(1)) - W and K_(2) = (hc)/(lamda_(2)) - W`
Hence, `K_(1) - (K_(2))/(3) = hc ((1)/(lamda_(1)) - (1)/(3lamda_(2)))- W + (W)/(3)`
`= hc.0 - (2W)/(3) [ :' lamda_(1) = 3 lamda_(2)]`
`= -(2W)/(3) lt 0`
`:. K_(1) lt (K_(2))/(3)`