Question

wavelengths belonging to Balmer series lying in the range of $450$ nm to $750$ nm were used to eject photoelectrons from a metal surface whose work function is $2.0eV$, Find (in eV) the maximum kinetic energy of the emitted photoelectrons. (Take hc=$1242eV$ nm.)

Answer 1

For maximum kinetic energy we need to incident the $highest$ energy photon on the surface so we need to do calculation for $450nm$ because being $lowest$ $wavelength$ it will have $greatest$ $energy.$

Energy of incident photon $E=\frac{hc}{\lambda }=\frac{1242}{450}eV=2.76eV$

Out of this $2.76eV$, $2eV$ will be consumed in dealing with $Work$ $function$ so maximum kinetic energy $K{E}_{max}=(2.76-2)eV=0.76eV$