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 os ejected electron (Given hc = 1242 eV nm)`

Explanation

KEY CONCEPT :
The wavelength `lambda` of photo for different lines of bamber series is given by
`(hc)/(lambda)= 13.6 [(1)/(2^(2)) - (1)/(n^(2))] eV where n = 3,4,5`
Using above relation , we get the value of `lambda = 657 nm , 487 nm between 450nm and 700 nm ` is smaller then `657 nm` , electron of max K.E will be emitted for photon corresponding to wavelength `487 nm` with
`(K.E) = (hc)/(lambda) - W = ((1242)/(487) - 2) = 0.55 eV`