Question

A metal plate is exposed to light of wavelength $\lambda$. It is observed that electrons are ejected from the surface of the plate. When a retarding uniform electric field $E$ is imposed, 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 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. ${\lambda }_0=\lambda -\frac{eEd}{hc}$

Answer 1

The correct option is B ${\lambda }_0={(\frac{1}{\lambda }-\frac{eED}{hc})}^{-1}$

We know the photoelectric equation as
$\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}=K.E_{max}$
$\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}=qEd$
where
$q=\text{charge of an electron}E=\text{eleectric field set up}d=\text{distance travelled by the electron}$
$\frac{1}{{\lambda }_0}=\frac{1}{\lambda }-\frac{eEd}{hc}$
${\lambda }_0=(\frac{1}{\lambda }-\frac{eEd}{hc}{)}^{-1}$