Wavelengths belonging to Balmer series for hydrogen atom lying in the range of 450 nm to 750 nm were used to eject photoelectrons from a metal surface whose work-function is 2.0 eV. Find (in eV) the maximum kinetic energy of the emitted photoelectrons.

Answer

`0.55 eV`

Explanation

Wavelengths corresponding to minimum wavelength `(lambda_("min"))` or maximum kinetic energy. `(lambda_("min"))` belonging to Balmer series and lying in the given range (450 nm to 750 nm) corresponding to transition from (n=4 to n=2). Here,
`E_(4)= - 13.6/((4)^(2))= -0.85 eV`
`E_(2)=13.6/((2)^(2))= -3.4 eV`
`:. DeltaE=E_(4)-E_(2)=2.55 eV`
`K_("max'")=` Energy of photon - work function
`=2.55-2.0=0.55 eV`