Question

Monochromatic light of wavelength 396 nm is incident on the surface of a metal whose work function is 1.125 eV. Calculate:

1. the energy of an incident photon in eV.
2. the maximum kinetic energy of photoelectrons in eV.

Answer 1

(i) Energy of proton = `"hc"/lambda`

`= (6.6 xx 10^-34 xx 3 xx 10^8)/(396 xx 10^-9)`

= 5 × 10^-19 J

= 3.125 eV

(ii) `"E"_"max" = "hc"/lambda - "W"`

= 3.125 - 1.125

= 2 eV.