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(D) The electric field is given by the sum of two waves with angular frequencies $\omega_1 = 6 	imes 10^{15} \, rad/s$ and $\omega_2 = 9 	imes 10^{1

Vedclass · Accepted Answer

$3.42$

Vedclass · Answer

(D) The electric field is given by the sum of two waves with angular frequencies $\omega_1 = 6 	imes 10^{15} \, rad/s$ and $\omega_2 = 9 	imes 10^{15} \, rad/s$.To find the maximum kinetic energy,we consider the photon with the higher frequency,as $K_{max} = hf - \phi$.The frequency $f$ is given by $f = \frac{\omega}{2\pi}$.For the higher frequency component,$f = \frac{9 	imes 10^{15}}{2\pi} \, Hz$.The energy of the photon is $E_{photon} = hf = (4.14 	imes 10^{-15} \, eVs) 	imes \left( \frac{9 	imes 10^{15}}{2 	imes 3.14159} ight) \, Hz$.$E_{photon} = \frac{37.26}{6.283} \approx 5.93 \, eV$.The maximum kinetic energy is $K_{max} = E_{photon} - \phi = 5.93 \, eV - 2.50 \, eV = 3.43 \, eV$.The closest option provided is $3.42 \, eV$.

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