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The work function of caesium is \(2.14\mathit{eV.}\) Find
The wavelength of the incident light if the photo current is brought to zero by a stopping potential of   \(0.06V\) .


The photo current is brought to zero, this happens when the maximum kinetic energy of photons becomes equal to the potential energy \(v_o\)  by the stopping potential \(v_o\)  i.e.
\(h=6.62\times 10^{-34} \;Js\)
\(e=1.6\times 10^{-19}C\)
\(ev_o=\mathit{hv}-\phi_o=\frac{\mathit{hc}}{\lambda }-\phi_o\)
\(\lambda =\frac{\mathit{hc}}{ev_o+\phi_o}\)
\(=\frac{6.62\times 10^{-34}\times 3\times 10^8}{\left(1.6\times 10^{-19}\times 0.60\right)+\left(2.14\times 1.6\times 10^{-19}\right)}\)
\(=4.54\times 10^{-7}\text{ m }\)
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