Q 12-11-076JEE MainJEE Main 2024 (31 Jan, Shift 1)Medium
When a metal surface is illuminated by light of wavelength $\lambda$, the stopping potential is $8\ \text{V}$. When the same surface is illuminated by light of wavelength $3\lambda$, the stopping potential is $2\ \text{V}$. The threshold wavelength for this surface is:
Answer: (C) $9\lambda$
In eV: $\dfrac{hc}{\lambda} = 8 + \phi$ and $\dfrac{hc}{3\lambda} = 2 + \phi$.
Subtracting: $\dfrac23\dfrac{hc}{\lambda} = 6 \Rightarrow \dfrac{hc}{\lambda} = 9\ \text{eV}$, so $\phi = 1\ \text{eV}$.
$$\lambda_0 = \frac{hc}{\phi} = 9\lambda$$
Solution by Sreeraj P, M.Sc Physics