(i) How does one explain the emission of electron from a photosensitive surface with te help of Einstein's photoelectric equation ?
(ii) The work function of the following metals is given : `Na = 2.75 eV, K = 2.3 eV, Mo = 4.17 eV` and `Ni = 5.15 eV`. Which of these metals will not cause photoelectric emission for radiation of wavelength `3300 Å` from a laser source placed 1 m aways from these metals ? What happens if the laser source is brought nearer and placed 50 cm away ?

Explanation

(i) When a proton of energy E = hv is incident on a metal surface, an energy `phi` is used up in liberating the `e^(-)`and the balance energy appears as `K.E. K_("max")` of the electron.
`rArr hv = phi_(o) + K_("max")"………"(i)`
If we go on reducing v, then for certain frequency `v_(o)` (threshold frequency) `K_("max")` becomes equal to zero.
Then, `hv_(o) - phi_(o)".........."`
Combing (i) and (ii)
`hv = hv_(o) + K_("max")`
`h (v - v_(o)) = K_("max") = 1/2 mv_("max")^(2)`
Which is Einstein's Photo-electric equation.
(ii) We know, `(hc)/(lambda) = phi + KE`
Emission will take place for `(hc)/(lambda) gt phi`
But, `(hc)/(lambda) (6.6 xx 10^(-34)xx 3 xx 10^(8))/(3300 xx 10^(-10)) = (19.8 xx 10^(-16))/(3.3 xx 10^(3)) = 6 xx 10^(-19) J = 3.75 eV`
`:.` No emission for Mo and Ni as their work function is high.
Work function depends on the frequency of the incident light. So, the distance of the laser is not going to affect the photoelectric emission.