Question

Ligth of wavelength `4000 Å` is incident on a metal plate whose work function is `2 eV`. What is maximum kinetic enegy of emitted photoelectron ?
A. `0.5 eV`
B. `1.1 eV`
C. `2.0 eV`
D. `1.5 eV`

Answer 1

Correct Answer - B
(b) If the maximum kinetic energy of photo electrons emitted from metal surface is `E_(k)` and `W` is the work-function of metal then
`E_(k) = (hc)/(lambda) - W`
where `hv` is the energy of photon absorbed by the electron in metal.
`: E_(k) = (hc)/(lambda) - W`
where `v = (c )/(lambda)`
Putting the numerical values, we have
`E_(k) = [(6.6 xx 10^(-34) xx 3 xx 10^(8))/(4000 xx 10^(-10) xx 1.6 xx 10^(-19)) - 2] eV`
`E_(k) = 3.1 - 2 = 1.1 eV`