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(C) The given electric field equation is $\varepsilon = (100 	ext{ V/m}) [\sin(\omega_1 t) + \sin(\omega_2 t)]$, where $\omega_1 = 5 	imes 10^{15} \

Vedclass · Accepted Answer

$3.27$

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(C) The given electric field equation is $\varepsilon = (100 	ext{ V/m}) [\sin(\omega_1 t) + \sin(\omega_2 t)]$, where $\omega_1 = 5 	imes 10^{15} 	ext{ rad/s}$ and $\omega_2 = 8 	imes 10^{15} 	ext{ rad/s}$.To find the maximum kinetic energy, we must consider the photon with the highest frequency, as $K.E._{\max} = h
u - \phi$.The frequency $
u$ is given by $
u = \frac{\omega}{2\pi}$. For the higher frequency component, $
u = \frac{8 	imes 10^{15}}{2\pi} 	ext{ Hz}$.The energy of the photon $E = h
u = \frac{6.626 	imes 10^{-34} 	imes 8 	imes 10^{15}}{2 	imes 3.1416} 	ext{ Joules}$.Converting this to $	ext{eV}$ by dividing by $1.6 	imes 10^{-19} 	ext{ C}$:$E = \frac{6.626 	imes 10^{-34} 	imes 8 	imes 10^{15}}{2 	imes 3.1416 	imes 1.6 	imes 10^{-19}} \approx 5.27 	ext{ eV}$.The maximum kinetic energy is $K.E._{\max} = E - \phi = 5.27 	ext{ eV} - 2 	ext{ eV} = 3.27 	ext{ eV}$.

Light described at a place by the equation $\varepsilon = (100 \text{ V/m}) [\sin(5 \times 10^{15} \text{ s}^{-1})t + \sin(8 \times 10^{15} \text{ s}^{-1})t]$ falls on a metal surface having a work function of $2 \text{ eV}$. Calculate the maximum kinetic energy of the photoelectrons in $\text{eV}$.

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