Question
A light wave described by $E = 60sin\left( 3 \times 10^{15} \right)t + sin\left( 12 \times 10^{15} \right)$ t] (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 \times 10^{- 34}\text{ }J - s.\ \right.\ $ and $e = 1.6 \times 10^{- 19}C$ )
Options
Solution
$\ \omega_{1} = 3 \times 10^{15}rad/sec$
$$\begin{matrix} & \omega_{2} = 12 \times 10^{15}rad/sec \\ & \ \because\ v = \frac{\omega}{2\pi} \\ & \begin{matrix} E_{\text{photon~}} & \ = hv = 6.6 \times 10^{- 34} \times 1.91 \times 10^{15} \\ & \ = 1.26 \times 10^{- 18}\text{ }J \\ E_{\max} & \ = \frac{1.26 \times 10^{- 18}}{1.6 \times 10^{- 19}} \simeq 7.9eV \\ {\text{ }K}_{\max} = & E_{\max} - \phi_{0} \\ & \ = 7.9 - 2.8 \\ {\text{ }K}_{\max} & \ = 5.1eV \end{matrix} \end{matrix}$$
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