When radiation of wavelength $\lambda$ is used to illuminate a metallic surface, the stopping potential is V. When the same surface is illuminated with radiation of wavelength 3$\lambda$, the stopping potential is ${V \over 4}$. If the threshold wavelength for the metallic surface is n$\lambda$ then value of n will be __________.
Answer (integer)
9
Solution
${{hc} \over \lambda } - \phi$ = eV .....(i)
<br><br>${{hc} \over {3\lambda }} - \phi = {{eV} \over 4}$ .....(ii)
<br><br>From (i) and (ii),
<br><br>${{hc} \over {3\lambda }} - \phi =$ ${{hc} \over {4\lambda }} - {\phi \over 4}$
<br><br>$\Rightarrow$ $${{hc} \over \lambda }\left( {{1 \over 3} - {1 \over 4}} \right) = {{3\phi } \over 4}$$
<br><br>$\Rightarrow$ ${{hc} \over {9\lambda }} = \phi$
<br><br>Also, $\phi$ = ${{hc} \over {{\lambda _0}}}$
<br><br>$\therefore$ ${{hc} \over {9\lambda }} = {{hc} \over {{\lambda _0}}}$
<br><br>$\Rightarrow$ $\phi$ = 9$\lambda$
<br><br>So, n = 9
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Photoelectric Effect
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