The surface of a metal is illuminated with the light of $400 \mathrm{~nm}$. The kinetic energy of the…

The surface of a metal is illuminated with the light of $400 \mathrm{~nm}$. The kinetic energy of the ejected photoelectrons was found to be $1.68 \mathrm{eV}$. The work function of the metal is $(\mathrm{hc}=1240 \mathrm{eV} \mathrm{nm})$
  1. $3.09 \mathrm{eV}$
  2. $1.41 \mathrm{eV}$
  3. $151 \mathrm{eV}$
  4. $1.68 \mathrm{eV}$

Solution

$ \begin{aligned} & \frac{1}{2} \mathrm{mv}^2=\mathrm{eV}_{\mathrm{o}}=1.68 \mathrm{eV} \Rightarrow \mathrm{hv}=\frac{\mathrm{hc}}{\lambda}=\frac{1240 \mathrm{evnm}}{400 \mathrm{~nm}}=3.1 \mathrm{eV} \Rightarrow 3.1 \mathrm{eV}=\mathrm{W}_0+1.6 \mathrm{eV} \\ & \therefore \quad \mathrm{W}_0=1.42 \mathrm{eV} \end{aligned} $

Asked in: JEE Main 2009

Practice more Dual Nature of Matter and Radiation questions on Aicharya