Maximum kinetic energy of the released photoelectrons emitted from metallic sodium, when a light is incident on it, 0.73 eV. If the work function of sodium is 1.82eV, find the energy of the incident photon in eV. Find wavelength of incident light.
(`h = 6.63 xx 10^(-27)` erg. s, `1eV = 1.6 xx 10^(-12)`erg)

Explanation

From Einstein's photoelectric equation, `E_("max") = hf - W_(0)`, we get the energy of incident photon as,
`E = hf = E_("max") + W_(0) = 0.73 + 1.82 = 2.55eV`
Hence wavelength of incident light, `lamda = (c)/(f) = (c)/(E//h)`
`:. lamda = (hc)/(E) = (6.63 xx 10^(-27) xx 3 xx 10^(10))/(2.55 xx 1.6 xx 10^(-12)) cm`
`= (6.63 xx 10^(-27) xx 3 xx 10^(10) xx 10^(8))/(2.55 xx 1.6 xx 10^(-12)) Å = 4875 Å`