The maximum kinetic energy of a photoelectron | vedclass

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Question

(D) Given,work function $W_0 = 2.35 	ext{ eV}$. The electric component of the electromagnetic radiation is given by $E = a[1 + \cos(2 \pi f_1 t)] \co

Vedclass · Accepted Answer

$17.45$

Vedclass · Answer

(D) Given,work function $W_0 = 2.35 	ext{ eV}$. The electric component of the electromagnetic radiation is given by $E = a[1 + \cos(2 \pi f_1 t)] \cos(2 \pi f_2 t)$.Using the trigonometric identity $\cos A \cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)]$,we expand the expression:$E = a \cos(2 \pi f_2 t) + a \cos(2 \pi f_1 t) \cos(2 \pi f_2 t)$$E = a \cos(2 \pi f_2 t) + \frac{a}{2} \cos[2 \pi (f_1 + f_2) t] + \frac{a}{2} \cos[2 \pi (f_1 - f_2) t]$.The frequencies present in the radiation are $f_2$,$(f_1 + f_2)$,and $(f_1 - f_2)$.To obtain the maximum kinetic energy,we must use the photon with the maximum frequency,which is $f_{\max} = f_1 + f_2 = 3.6 	imes 10^{15} + 1.2 	imes 10^{15} = 4.8 	imes 10^{15} 	ext{ Hz}$.The energy of this photon is $E_{	ext{photon}} = h f_{\max} = (6.6 	imes 10^{-34} 	ext{ Js} 	imes 4.8 	imes 10^{15} 	ext{ Hz}) / (1.6 	imes 10^{-19} 	ext{ J/eV}) = 19.8 	ext{ eV}$.The maximum kinetic energy is $KE_{\max} = E_{	ext{photon}} - W_0 = 19.8 	ext{ eV} - 2.35 	ext{ eV} = 17.45 	ext{ eV}$.Thus,the correct option is $D$.

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