Photo Electric Effect

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Question

In a photo electric experiment, work function of a target metal is 1ev and wavelength of light used in the experiment is $3000 \overset{o}{A}$ Then stopping potential will be

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Solution

Energy of incident photon,
$\begin{array}{l} hν = \frac{hc}{λ} = \frac{12375}{3000} ev \\ = 4 .125 ev \end{array}$
$Now hν = ϕ + {ev}_{0} \Rightarrow V_{0} = (\frac{4 .125 - 1}{e}) ev = 3 .125 V$

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