A light wave described by E = 60 sin(3 times | vedclass

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(A) The given electric field equation is $E = 60 \sin(3 	imes 10^{15} t) + \sin(12 	imes 10^{15} t)$.This represents two light waves with angular fr

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$5.1$

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(A) The given electric field equation is $E = 60 \sin(3 	imes 10^{15} t) + \sin(12 	imes 10^{15} t)$.This represents two light waves with angular frequencies $\omega_1 = 3 	imes 10^{15} 	ext{ rad/s}$ and $\omega_2 = 12 	imes 10^{15} 	ext{ rad/s}$.The energy of a photon is given by $E_{ph} = \hbar \omega = \frac{h \omega}{2\pi}$.For the higher frequency component $\omega_2 = 12 	imes 10^{15} 	ext{ rad/s}$:$E_{ph} = \frac{6.63 	imes 10^{-34} 	imes 12 	imes 10^{15}}{2 	imes 3.14} 	ext{ J}$.$E_{ph} \approx 1.265 	imes 10^{-18} 	ext{ J}$.Converting this to electron-volts $(	ext{eV})$: $E_{ph} = \frac{1.265 	imes 10^{-18}}{1.6 	imes 10^{-19}} \approx 7.9 	ext{ eV}$.According to Einstein's photoelectric equation, $K_{max} = E_{ph} - \phi_0$.Given work function $\phi_0 = 2.8 	ext{ eV}$.$K_{max} = 7.9 	ext{ eV} - 2.8 	ext{ eV} = 5.1 	ext{ eV}$.

$A$ light wave described by $E = 60 \sin(3 \times 10^{15} t) + \sin(12 \times 10^{15} t)$ (in $SI$ units) falls on a metal surface of work function $2.8 \text{ eV}$. The maximum kinetic energy of the ejected photoelectron is (approximately) . . . . . . $\text{eV}$.

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