A light source of wavelength $\lambda$ illuminates a metal surface and electrons are ejected with maximum…

A light source of wavelength $\lambda$ illuminates a metal surface and electrons are ejected with maximum kinetic energy of 2 eV. If the same surface is illuminated by a light source of wavelength $\frac{\lambda}{2}$, then the maximum kinetic energy of ejected electrons will be (The work function of metal is 1 eV)
  1. 3 eV
  2. 2 eV
  3. 6 eV
  4. 5 eV

Solution

Einstein's photoelectric equation
$\begin{aligned}
& \mathrm{KE}=\frac{h c}{\lambda}=\phi_0 \\ & 2 \mathrm{eV}=\frac{h c}{\lambda}-1 \mathrm{eV} \\ & \frac{h c}{\lambda}=3 \mathrm{eV} \\ & \mathrm{KE}^{\prime}=\frac{h c}{(\lambda / 2)}-\phi_0=6 \mathrm{eV}-1 \mathrm{eV}
\end{aligned}$
$=5 \mathrm{eV}$

Asked in: JEE Main 2025 (22 Jan Shift 2)

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