Question

Radiation from hydrogen gas excited to first excited state is used for illuminating certain photoelectric plate. When the radiation from some unknown hydrogen-like gas excited to the same level is used to expose the same plate, it is found that the de Broglie wavelength of the fastest photoelectron has decreased $2.3$ times. It is given that the energy corresponding to the longest wavelength of the Lyman series of the unknown gas (13.6 eV). Find the work function of the photoelectric plate in eV. [Take $(2.3{)}^2=5.25$]

Answer 1

According to photoelectric law:

${KE}_{max}=\frac{p^2}{2m}=h\vartheta -Wusingde-Broglierelation:p=\frac{h}{\lambda }\frac{h^2}{2m{\lambda }^2}=h\vartheta -W\Rightarrow \lambda =\frac{h}{\sqrt{2m(h\vartheta -W)}}and{\lambda }^{\prime }=\frac{h}{\sqrt{2m(h{\vartheta }^{\prime }-W)}}=\frac{\lambda }{2.3}\to 2.3{\lambda }^{\prime }=\lambda \Rightarrow 5.25h\vartheta -5.25W=h{\vartheta }^{\prime }-W-\left(1\right)since,2.3{\lambda }^{\prime }=\lambda 2.3\frac{c}{{\vartheta }^{\prime }}=\frac{c}{\vartheta }\to 2.3\vartheta =\vartheta \therefore \left(1\right)becomes5.25h\vartheta -5.25W=2.3h\vartheta -W\to W=\frac{2.95}{4.25}W=0.69h\vartheta =9.44eV$