Question

A horizontal cesium plate (φ = 1.9 eV) is moved vertically downward at a constant speed v in a room full of radiation of wavelength 250 nm and above. What should be the minimum value of v so that the vertically-upward component of velocity is non-positive for each photoelectron?

Answer 1

Given:
Work function of the cesium plate, φ = 1.9 eV
Wavelength of radiation, λ = 250 nm
Energy of a photon,
$$\begin{gathered} E=\frac{hc}{\lambda }, \\ \text{w}\mathrm{here}h=\text{P}\mathrm{lanck}\text{'s}\mathrm{constant} \\ c=\mathrm{speed}\mathrm{of}\mathrm{light} \\ \therefore E=\frac{1240}{250}=4.96\mathrm{eV} \end{gathered}$$
From Einstein's photoelectric equation, kinetic energy of an electron,
$$\begin{gathered} K=E-\varphi \\ \Rightarrow K=\frac{hc}{\lambda }-\varphi \\ \left[\text{H}\mathrm{ere},h\text{is}\mathrm{Planck}\text{'s}\mathrm{constant}\mathrm{and}\mathrm{c}isthe\mathrm{speed}\mathrm{of}\mathrm{light}\right] \\ \Rightarrow K=4.96\mathrm{eV}-1.9\mathrm{eV} \\ =3.06\mathrm{eV}. \end{gathered}$$
For non-positive velocity of each photo electron, the velocity of a photoelectron should be equal to minimum velocity of the plate.
∴ Velocity of the photoelectron,
$$\begin{gathered} v=\sqrt{\frac{2K}{m}}\left(m=\mathrm{mass}\mathrm{of}\mathrm{electron}\right) \\ \therefore v=\sqrt{\frac{2\times 3.06\times 1.6\times {10}^{-19}}{9.1\times {10}^{-31}}} \\ =1.04\times {10}^6{\mathrm{ms}}^{-1} \end{gathered}$$