Photoelectric Example

The work function is

$eW_{0}=h\nu _{th}={\frac {hc}{\lambda }}={\frac {12400{\text{ eV}}\cdot \mathrm {\AA} }{6000\,\mathrm {\AA} }}=2.07{\text{ eV}}$

The photoelectric equation then gives

$eV_{s}=h\nu -eW_{0}={\frac {hc}{\lambda }}-eW_{0}$

$2.5{\text{ eV}}={\frac {12400{\text{ eV}}\cdot \mathrm {\AA} }{\lambda }}-2.07{\text{ eV}}\Rightarrow \lambda =2713\,\mathrm {\AA}$