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

Here, `phi_(0)=2eV, (hc)/e=1242eV-nm`,
The wavelength of incident radiation lies between 450nm to 700nm of Balmer series. It arises when transition takes place from `n_(2)=4` to `n_(1)=2` in case of hydrogen atom. Fot the transition, `n_(1)=2 and n_(2)=n`
energy emitted is `DeltaE=13.6(1/(2^(2))-1/(n^(2)))=(hc)/(elambda)=1242/lambda or lambda=(1242xx4n^(2))/(13.6(n^(2)-4))nm`
When n=4, then `lambda=(1242xx4xx4n^(2))/(13.6(4^(2)-4))=487.05nm`
Maximum K.E. of emitted photoelectron is, `E_(max)=(hc)/(elambda)-phi_(0) (i n eV) =1242/487.05-2=2.55-2=0.55eV`