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## QuestionPhysicsClass 12

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

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