A surface has light of wavelength ${\mathrm{\lambda }}_1=550 \mathrm{nm}$ incident on it, causing the ejection of photoelectrons for which the stopping potential is ${\mathrm{V}}_{{\mathrm{S}}_1}=0.19 \mathrm{V}$. If the radiation of wavelength ${\mathrm{\lambda }}_2=190 \mathrm{nm}$ is now incident on the surface, 

(1) For the first wavelength: ${\mathrm{eV}}_{{\mathrm{s}}_1}=\mathrm{hf}-\mathrm{W}$                        …(i)

     W = work function

     For the second surface: ${\mathrm{eV}}_{{\mathrm{s}}_2}={\mathrm{hf}}_2-\mathrm{W}$                       …(ii)

     Subtracting, we get ${\mathrm{V}}_{{\mathrm{s}}_2}-{\mathrm{V}}_{{\mathrm{s}}_1}=\frac{\mathrm{h}}{\mathrm{e}}\left({\mathrm{f}}_2-{\mathrm{f}}_1\right)$

     $\begin{array}{l}\mathrm{or} {\mathrm{V}}_{{\mathrm{s}}_2}={\mathrm{V}}_{{\mathrm{s}}_1}+\frac{\mathrm{hc}}{\mathrm{e}}\left(\frac{1}{{\mathrm{\lambda }}_2}-\frac{1}{{\mathrm{\lambda }}_1}\right)={\mathrm{V}}_{{\mathrm{s}}_1}+\frac{\mathrm{hc}}{\mathrm{e}}\left(\frac{{\mathrm{\lambda }}_1-{\mathrm{\lambda }}_2}{{\mathrm{\lambda }}_1{\mathrm{\lambda }}_2}\right)\\ =0.19+1240\left(\frac{550-190}{190\times 550}\right)=4.46\mathrm{V}\end{array}$

(2) From Eq. (i): $\mathrm{W}=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_1}-{\mathrm{eV}}_{{\mathrm{s}}_1}$

     Work function (in eV),

           $\mathrm{W}=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_1}-{\mathrm{eV}}_{{\mathrm{s}}_1}=\frac{1240}{550}-0.19=2.06 \mathrm{eV}$

(3) Threshold frequency,

            ${\mathrm{f}}_0=\frac{\mathrm{W}}{\mathrm{h}}=\frac{2.06\times 1.6\times {10}^{-19}}{6.63\times {10}^{-34}}\approx 5\times {10}^{14}\mathrm{Hz}$