Light of wavelength `2000 Å` falls on an aluminium surface . In aluminium `4.2 e V` of energy is required to remove an electron from its surface. What is the kinetic energy , in electron volt of (a) the fastest and (b) the slowest emitted photo-electron . ( c) What is the stopping potential ? (d) What is the cut - off wavelength for aluminum? (Plank's constant `h = 6.6 xx 10^(-34) J-s` and speed of light `c = 3 xx 10^(8) m s^(-1).

Explanation

`lambda =2000 Å =200 nm, phi =4.2 eV`
Energy of incident photon, `E =(1242)/(200) =6.21 eV`
(a) `E=phi + K_(max)`
`6.21 = 4.2 + K_(max) implies K_(max) =2 eV`
(b) The kinetic energy of emitted electrons vary from zero to `K_(max)`
`K_(min =0`
(c) `K_(max) =eV_(s) implies 2=V_(s)`
`V_(s) =2V`
(d) Cut-off wavelength `lambda_(0) =(1242)/(phi(eV)) = (1242)/(4.2) =300 nm`