The photons from the Balmer series in Hydrogen spectrum having wavelength between `450 nm` to `700 nm` are incident on a metal surface of work function `2 eV` Find the maximum kinetic energy of ejected electron (Given hc = 1242 eV nm)`

Explanation

Balmer series
`lambda_(32) = (12375)/(E_3 - E_2) = (12375)/((13.6)((1)/(4)- (1)/(9))`
=6551 Å
= 655.1 nm
`lambda_(42) = (12375)/(E_4 -E_2) = (12375)/((13.6)((1)/(4) - (1)/(16)))`
= 4853 Å
=433.3 nm
First two lie in the given range. of these `lambda_(42)`
corresponds to more energy
`E = E_4 - E_2 = (13.6)((1)/(4) - (1)/(16))`
=2.55 eV
`K_(max) = E -W = (2.55 - 2.0) eV`
= 0.55 eV.