Question

Find the maximum kinetic energy of the photoelectrons ejected when light of wavelength 350 nm is incident on a cesium surface. Work function of cesium = 1.9 eV

Answer 1

Given:
Wavelength of light, λ = 350 nm = 350 × 10⁻⁹ m
Work-function of cesium, ϕ = 1.9 eV
From Einstein's photoelectric equation,
$$\begin{gathered} E=\varphi +\mathrm{Kinetic}\mathrm{energy}\mathrm{of}\mathrm{electron} \\ \Rightarrow \mathrm{K}\mathrm{E}=E-\varphi \\ \Rightarrow \mathrm{K}\mathrm{E}=\frac{hc}{\lambda }-\varphi , \\ \text{w}\mathrm{here}\lambda =\mathrm{wavelength}\mathrm{of}\mathrm{light} \\ h=\text{P}\mathrm{lan}\text{ck}\text{'}\mathrm{s}\mathrm{constant} \end{gathered}$$
Maximum kinetic energy of electrons,
$$\begin{gathered} E_{\max }=\frac{hc}{\lambda }-\varphi \\ {E}_{\max }=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{350\times {10}^{-9}\times 1.6\times {10}^{-19}}-1.9 \\ {E}_{\max }=\frac{6.63\times 3\times {10}^2}{350\times 1.6}-1.9 \\ {E}_{\max }=1.65\mathrm{eV}=1.6\mathrm{eV} \end{gathered}$$