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(A) Given: Power of source $P = 1 \,W$, distance $d = 1 \,m$, radius of exposed area $r = 1 \,Å = 10^{-10} \,m$, work function $\phi = 5 \,eV = 5 	im

Vedclass · Accepted Answer

$320$

Vedclass · Answer

(A) Given: Power of source $P = 1 \,W$, distance $d = 1 \,m$, radius of exposed area $r = 1 \,Å = 10^{-10} \,m$, work function $\phi = 5 \,eV = 5 	imes 1.6 	imes 10^{-19} \,J$.Intensity $I$ at distance $d$ is $I = \frac{P}{4 \pi d^2} = \frac{1}{4 \pi (1)^2} = \frac{1}{4 \pi} \,W/m^2$.Power absorbed by the circular area $A = \pi r^2$ is $P_{abs} = I 	imes A = \frac{1}{4 \pi} 	imes \pi (10^{-10})^2 = \frac{10^{-20}}{4} \,W$.The time $t$ required to accumulate energy equal to the work function is $t = \frac{\phi}{P_{abs}}$.Substituting the values: $t = \frac{5 	imes 1.6 	imes 10^{-19}}{10^{-20} / 4} = \frac{8 	imes 10^{-19}}{0.25 	imes 10^{-20}} = 320 \,s$.

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