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
$E = \dfrac{1240}{300} \approx 4.13$ eV. $K_{max} = 4.13 - 2.14 \approx 2.0$ eV.
Solution by Sreeraj P, M.Sc Physics