Question

A metal plate is exposed to light with wavelength $\lambda$. It is observed that electrons are ejected from the surface of the plate. When a retarding uniform electric field $E$ is applied, no electron can move away from the plate farther than a certain distance $d$. Then the threshold wavelength ${\lambda }_0$ for the material of plate is ($e$ is the electronic charge, $h$ is the Plank’s constant and $c$ is the speed of light.)

• A. ${\lambda }_0=(\frac{1}{\lambda }-\frac{hc}{eEd}{)}^{-1}$
• B. ${\lambda }_0=(\frac{1}{\lambda }-\frac{eEd}{hc}{)}^{-1}$
• C. ${\lambda }_0=\lambda -\frac{hc}{eEd}$
• D. ${\lambda }_0=(\lambda -\frac{eEd}{hc}{)}^{-1}$

Answer 1

The correct option is B

${\lambda }_0=(\frac{1}{\lambda }-\frac{eEd}{hc}{)}^{-1}$

The maximum kinetic energy of the electron immediately upon ejection is the difference between the energy of the incident light and the threshold energy

$K=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}$ ....(i)

The kinetic energy of ejected electron is converted to electrostatic potential energy i.e. $\Delta U=eEd$ Therefore, $K=Eed$ and putting this in eq(i)

${\lambda }_0=(\frac{1}{\lambda }-\frac{eEd}{hc}{)}^{-1}$