$\begin{aligned}
& \phi=2.14 \
& V_S=2 \mathrm{V}
\end{aligned}Usingphotoelectricequation.\begin{aligned}
& \frac{h c}{\lambda}=2.14+2=4.14 \mathrm{eV} \
& \lambda=\frac{1242}{4.14}=300 \mathrm{nm}
\end{aligned}$
JEE Main 2025 — Physics Modern Physics
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 =1242eVnm where h is the Planck's constant and c is the speed of light in vaccum.)
Held on 23 Jan 2025 · Verified 6 Jul 2026.
300 nm
400 nm
600 nm
200 nm
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