A photoelectric surface is illuminated successively by monochromatic light of 'wavelength ' $\lambda$ ' and…
A photoelectric surface is illuminated successively by monochromatic light of 'wavelength ' $\lambda$ ' and $\left(\frac{\lambda}{2}\right)$. If the maximum kinetic energy of the emitted photoelectrons in the first case is one-fourth that in the second case, the work function of the surface of the material is ( $\mathrm{c}=$ speed of light, $\mathrm{h}=$ Planck's constant $)$
$\frac{2 \mathrm{hc}}{\lambda}$
$\frac{\mathrm{hc}}{\lambda}$
$\frac{2 \mathrm{hc}}{3 \lambda}$
$\frac{\mathrm{hc}}{3 \lambda}$
Solution
$\begin{aligned}
& E_1=\frac{h c}{\lambda}-\phi ...(i)\\
& E_2=\frac{h c}{\lambda / 2}-\phi=\frac{2 h c}{\lambda}-\phi \ldots \ldots \text { (ii) } \\
& \text { Given, } E_1=\frac{E_2}{4} \\
& 4 E_1=E_2 ...(iii)\\
& \text { From (ii) and (iii), } \\
& 4 E_1=\frac{2 h c}{\lambda}-\phi \\
& 4\left(\frac{h c}{\lambda}-\phi\right)=\frac{2 h c}{\lambda}-\phi
\end{aligned}$
[Substituting from (i)]
$\begin{array}{ll}\therefore & \frac{4 h c}{\lambda}-4 \phi=\frac{2 h c}{\lambda}-\phi \\ \therefore & 3 \phi=\frac{2 h c}{\lambda} \\ \therefore & \phi=\frac{2 h c}{3 \lambda}\end{array}$
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