• # question_answer When a metallic surface is illuminated with monochromatic light of wavelength $\lambda ,$ the stopping potential is $5\,{{V}_{0}}.$ When the same surface is illuminated with light of wavelength $3\lambda ,$ the stopping potential is ${{V}_{0}}.$ Then the work function of the metallic surface is: A) $\frac{hc}{6\lambda }$ B) $\frac{hc}{5\lambda }$ C) $\frac{hc}{4\lambda }$ D) $\frac{2\,hc}{4\lambda }$

Solution :

 $\frac{hc}{\lambda }=5\,\,e{{V}_{0}}+\phi$ $\frac{hc}{3\lambda }=\,\,e{{V}_{0}}+\phi$    $\Rightarrow$   $\frac{2hc}{3\lambda }=\,\,4e{{V}_{0}}$ $\Rightarrow$   $\phi =\frac{hc}{6\lambda }$    Ans.

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