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Question and Answer

The work function of cesium is \(2.14\mathit{eV}\) . Find
The threshold frequency for cesium.

Answer

For the cut-off or threshold frequency, the energy \(\mathit{hv}\)  of the incident
radiation must be equal to work function \(\varphi _O\) , so that
\(hv_o=\varphi _o\)
\(v_O=\frac{\varphi _o} h\)
\(\varphi _O=2.14\mathit{eV}=2.14\times 1.6\times 10^{-19}J\)
\(h=6.62\times 10^{-34}\mathit{Js}\)
\(v_o=\frac{2.14\times 1.6\times 10^{-19}}{6.62\times 10^{-34}}=5.16\times 10^{14}\) Hz
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