In photoelectric effect an EM-wave is incident on a metal surface and electrons are ejected from the surface…

In photoelectric effect an EM-wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.14 eV and stopping potential is 2 V, what is the wavelength of the EM-wave ?
(Given hc $=1242 \mathrm{eVnm}$ where h is the Planck's constant and c is the speed of light in vacuum.)
  1. 300 nm
  2. 400 nm
  3. 600 nm
  4. 200 nm

Solution

$\begin{aligned}
& \phi=2.14 \\ & V_S=2 \mathrm{~V}
\end{aligned}$
Using photoelectric equation.
$\begin{aligned}
& \frac{h c}{\lambda}=2.14+2=4.14 \mathrm{eV} \\ & \lambda=\frac{1242}{4.14}=300 \mathrm{~nm}
\end{aligned}$

Asked in: JEE Main 2025 (23 Jan Shift 2)

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