Q67 / 75 · JEE Main 2025 (22 Jan Shift 2) · Physics
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 )
- A3 eV
- B2 eV
- C6 eV
- Correct answerD5 eV
Solution
Einstein's photoelectric equation
$\begin{aligned}
& \mathrm{KE}=\frac{h c}{\lambda}=\phi_0 \\
& 2 \mathrm{eV}=\frac{h c}{\lambda}-1 \mathrm{eV} \\
& \frac{h c}{\lambda}=3 \mathrm{eV} \\
& \mathrm{KE}^{\prime}=\frac{h c}{(\lambda / 2)}-\phi_0=6 \mathrm{eV}-1 \mathrm{eV}
\end{aligned}$
$=5 \mathrm{eV}$
Hence, the correct answer is option: 'D'.