Einstein's photoelectric equation $\begin{aligned}
& \mathrm{KE}=\frac{h c}{\lambda}=\phi_0 \
& 2 \mathrm{eV}=\frac{h c}{\lambda}-1 \mathrm{eV} \
& \frac{h c}{\lambda}=3 \mathrm{eV} \
& \mathrm{KE}^{\prime}=\frac{h c}{(\lambda / 2)}-\phi_0=6 \mathrm{eV}-1 \mathrm{eV}
\end{aligned}=5 \mathrm{eV}$
A light source of wavelength λ illuminates a metal surface and electrons are ejected with maximum kinetic energy of 2 eV . If the same surface is…
JEE Main 2025 — Physics Modern Physics
A light source of wavelength λ 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 2λ, then the maximum kinetic energy of ejected electrons will be (The work function of metal is 1 eV )
Held on 22 Jan 2025 · Verified 6 Jul 2026.
Options
- A
3 eV
- B
2 eV
- C
6 eV
- D
5 eV
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