The electric field component of an EM radiati | vedclass

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(A, B) The given electric field is $E = a \cos \omega_{0} t + \frac{a}{2} [\sin(\omega + \omega_{0})t + \sin(\omega - \omega_{0})t]$.This indicates th

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The stopping potential vs. frequency graph will be a straight line.

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(A, B) The given electric field is $E = a \cos \omega_{0} t + \frac{a}{2} [\sin(\omega + \omega_{0})t + \sin(\omega - \omega_{0})t]$.This indicates the presence of three frequencies: $f_{0} = \frac{\omega_{0}}{2\pi}$, $f_{1} = \frac{\omega + \omega_{0}}{2\pi}$, and $f_{2} = \frac{|\omega - \omega_{0}|}{2\pi}$.The maximum energy corresponds to the highest frequency $f_{1} = \frac{6 	imes 10^{15}}{2\pi} 	ext{ Hz}$.The energy of this photon is $E_{max} = h f_{1} = \frac{6.62 	imes 10^{-34} 	imes 6 	imes 10^{15}}{2 	imes 3.14 	imes 1.6 	imes 10^{-19}} 	ext{ eV} \approx 3.95 	ext{ eV}$.Given stopping potential $V_{s} = 2 	ext{ V}$, so $KE_{max} = 2 	ext{ eV}$.Using $KE_{max} = E_{max} - \phi$, we get $2 	ext{ eV} = 3.95 	ext{ eV} - \phi$, so $\phi = 1.95 	ext{ eV}$.For frequency $\omega = 10^{15} 	ext{ s}^{-1}$, energy $E = h(\frac{\omega}{2\pi}) \approx 0.66 	ext{ eV}$.Since $E Einstein's photoelectric equation $eV_{s} = hf - \phi$ represents a straight line graph between $V_{s}$ and $f$.

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