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 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.

\(h=6.62\times 10^{-34} \;Js\)

\(e=1.6\times 10^{-19}C\)

\(ev_o=\mathit{hv}-\phi_o=\frac{\mathit{hc}}{\lambda }-\phi_o\)

\(\lambda =\frac{\mathit{hc}}{ev_o+\phi_o}\)

\(=\frac{6.62\times 10^{-34}\times 3\times 10^8}{\left(1.6\times 10^{-19}\times 0.60\right)+\left(2.14\times 1.6\times 10^{-19}\right)}\)

\(=4.54\times 10^{-7}\text{ m }\)

\(h=6.62\times 10^{-34} \;Js\)

\(e=1.6\times 10^{-19}C\)

\(ev_o=\mathit{hv}-\phi_o=\frac{\mathit{hc}}{\lambda }-\phi_o\)

\(\lambda =\frac{\mathit{hc}}{ev_o+\phi_o}\)

\(=\frac{6.62\times 10^{-34}\times 3\times 10^8}{\left(1.6\times 10^{-19}\times 0.60\right)+\left(2.14\times 1.6\times 10^{-19}\right)}\)

\(=4.54\times 10^{-7}\text{ m }\)

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