Q 12-11-021NEETJEE MainEasy
Light of wavelength $300$ nm falls on a metal of work function $2.14$ eV. The maximum kinetic energy of the photoelectrons is ($hc = 1240$ eV nm)
Answer: (A) $2.0$ eV
Concept. $K_{\max} = \dfrac{hc}{\lambda} - \phi$.
Step 1. Photon energy. $\dfrac{1240}{300} \approx 4.13$ eV.
Step 2. KE. $4.13 - 2.14 \approx 2.0$ eV.
Answer. $2.0$ eV (Option A).
Solution by Sreeraj P, M.Sc Physics