Q 12-11-132JEE MainJEE Main 2021 (25 Jul, Shift 2)Medium
When radiation of wavelength $\lambda$ is incident on a metallic surface, the stopping potential of ejected photoelectrons is $4.8$ V. If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes $1.6$ V. The threshold wavelength of the metal is:
Answer: (B) $4\lambda$
$\dfrac{hc}{\lambda} - \phi = 4.8$ eV and $\dfrac{hc}{2\lambda} - \phi = 1.6$ eV.
Subtracting: $\dfrac{hc}{2\lambda} = 3.2$ eV, so $\dfrac{hc}{\lambda} = 6.4$ eV and $\phi = 1.6$ eV $= \dfrac{1}{4}\cdot\dfrac{hc}{\lambda}$.
$\lambda_0 = \dfrac{hc}{\phi} = 4\lambda$.
Solution by Sreeraj P, M.Sc Physics