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-third 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.)
$\frac{2 \mathrm{hc}}{\lambda}$
$\frac{\mathrm{hc}}{2 \lambda}$
$\frac{\text { hc }}{\lambda}$
$\frac{\text { hc }}{3 \lambda}$
Solution
The correct option is (B).
Concept: Einstien's Photoelectric effect theory
The maximum kinetic energy $\mathrm{K}$ of the photoelectric electrons is related to the incident wavelength $\lambda$ and work function $\phi$ of the metal as follows:
$\mathrm{K}=\frac{\mathrm{hc}}{\lambda}-\phi$
For the incident wavelength $\lambda$, the equation is $\mathrm{K}=\frac{\mathrm{hc}}{\lambda}-\phi$
And for the incident wavelength $\frac{\lambda}{2}$, the equation is $3 K=\frac{\mathrm{hc}}{\frac{\lambda}{2}}-\phi$
Taking the ratio of two equations:
$\frac{1}{3}=\frac{\left(\frac{\mathrm{hc}}{\lambda}-\phi\right)}{\left(\frac{\mathrm{hc}}{\frac{\lambda}{2}}-\phi\right)}$
On solving, we get $\phi=\frac{\mathrm{hc}}{2 \lambda}$
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