Q 12-11-091JEE MainJEE Main 2023 (11 Apr, Shift 1)Medium
A metallic surface is illuminated with radiation of wavelength $\lambda$, the stopping potential is $V_0$. If the same surface is illuminated with radiation of wavelength $2\lambda$, the stopping potential becomes $\dfrac{V_0}4$. The threshold wavelength for this metallic surface will be
Answer: (A) $3\lambda$
$eV_0=\dfrac{hc}{\lambda}-\phi$ and $\dfrac{eV_0}{4}=\dfrac{hc}{2\lambda}-\phi$.
Multiply the second by 4 and subtract: $3\phi=\dfrac{hc}{\lambda}$, so $\phi=\dfrac{hc}{3\lambda}$ and $\lambda_0=3\lambda$.
Solution by Sreeraj P, M.Sc Physics