Question

The work function of a surface of a photosensitive material is 6.2eV. The wavelength of the incident radiation for which the stopping potential is 5V lies in the

1-X-Ray region

2-Ultraviolet region

3-Visible region

4-Infrared region

Answer 1

$$\begin{gathered} Energyoftheemittedphotoelectronisgivenby: \\ {E}_e=E-W \\ whereEistheenergyofincidentphotonWistheworkfucntion \\ Stoppoinpotentialoftheemittedphotoelctronintermsoftheenergyisgivenby: \\ {E}_e=eV \\ {E}_e=5eVolt \\ W=6.2evolt \\ E=5+6.2=11.2evolt \\ E=\frac{hc}{\lambda }=11.2evolt \\ \lambda =\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{11.2\times 1.6\times {10}^{-19}} \\ \lambda =1.1\times {10}^{-7}m \\ Sothisisvisiblelight \end{gathered}$$