A light wave described by $E=60\left[\sin \left(3 imes 10^{15} ight) t+\sin \left(12 imes 10^{1

Question

A light wave described by $E=60\left[\sin \left(3 	imes 10^{15}ight) t+\sin \left(12 	imes 10^{15}ight) tight]$ (in SI units) falls on a metal surface of work function 2.8 eV . The maximum kinetic energy of ejected photoelectron is (approximately) $\_\_\_\_$ eV. $\left(h=6.6 	imes 10^{-34}ight.$ J.s. and $\left.e=1.6 	imes 10^{-19} \mathrm{C}ight)$

Professor PC Thomas Classes · Answer

The maximum kinetic energy is determined by the highest frequency component of the incident light. The highest angular frequency is $\omega = 12 	imes 10^{15} 	ext{ rad/s}$. The energy of this photon is $E = \hbar \omega = \frac{h \omega}{2\pi} = \frac{6.6 	imes 10^{-34} 	imes 12 	imes 10^{15}}{2 	imes \pi} \approx 12.6 	imes 10^{-19} 	ext{ J}$. Converting to eV: $E \approx \frac{12.6 	imes 10^{-19}}{1.6 	imes 10^{-19}} \approx 7.87 	ext{ eV}$. Using Einstein's photoelectric equation, $K_{	ext{max}} = E - W = 7.87 	ext{ eV} - 2.8 	ext{ eV} = 5.07 	ext{ eV} \approx 5.1 	ext{ eV}$.