Question

Solve this:
7. The work function of a surface of a photosensitive material is 6.2 eV. The wavelength of the incident radiation for which the stopping potential is 5 V lies in the :-
(1) Infrared region
(2) X - ray region
(3) Ultraviolet region
(4) Visible region

Answer 1

$$\begin{gathered} Weknowthat \\ e{V}_S=E_{photon}-{\varphi }_0 \\ Given \\ {V}_S=5V \\ Workfunction{\varphi }_0=6.2eV \\ Note1eV=e\times Volts,whereeistheelectroniccharge \\ {E}_{photon}=\frac{1240}{\lambda \left(innm\right)}eV \\ e\left(5V\right)=\frac{1240}{\lambda \left(innm\right)}eV-6.2eV \\ 5eV=\frac{1240}{\lambda \left(innm\right)}eV-6.2eV \\ 5eV+6.2eV=\frac{1240}{\lambda \left(innm\right)}eV \\ \lambda =\frac{1240}{11.2}nm=110.71nm \end{gathered}$$
This wavelength lies in ultraviolet region