When photons of wavelength $λ_{1}$ are incident on an isolated sphere, the corresponding stopping potential is found to be V. When photons of wavelength $λ_{2}$ are used, the corresponding stopping potential was thrice that of the above value. If light of wavelength $λ_{3}$ is used then find the stopping Potential for this case:

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$\frac{h c}{e} [\frac{1}{λ_{3}} - \frac{1}{λ_{2}} - \frac{1}{λ_{1}}]$

$\frac{h c}{e} [\frac{1}{λ_{3}} + \frac{1}{λ_{2}} - \frac{1}{λ_{1}}]$

$\frac{h c}{e} [\frac{1}{λ_{3}} - \frac{3}{2 λ_{1}} + \frac{1}{2 λ_{2}}]$

$\frac{h c}{e} [\frac{1}{λ_{3}} + \frac{1}{2 λ_{2}} - \frac{1}{λ_{1}}]$

answer is D.

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Detailed Solution

From Einstein’s photoelectric equation, we have
$\frac{h c}{λ_{1}} = \frac{h c}{λ_{0}} + e V$ ……..(1)
$\frac{h c}{λ_{2}} = \frac{h c}{λ_{0}} + 3 e V$ ……. (2)
$\frac{h c}{λ_{3}} = \frac{h c}{λ_{0}} + e V^{'}$ ………. (3)
From equation (1)& (2)
$\begin{array}{l} \frac{3}{2 λ_{1}} - \frac{1}{2 λ_{2}} = \frac{1}{λ_{0}} \\ \Rightarrow \frac{h c}{λ_{1}} - h c [\frac{3}{2 λ_{1}} - \frac{1}{2 λ_{2}}] = e V^{'} \\ \Rightarrow \frac{h c}{e} [\frac{1}{λ_{3}} - \frac{3}{2 λ_{1}} + \frac{1}{2 λ_{2}}] = V^{'} \end{array}$