$(a)\;15s\qquad(b)\;17s\qquad(c)\;16s\qquad(d)\;18s$

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Answer : (b) 17s

Explanation :

If source radiates uniformly in all direction , the intensity I of light at a distance r is given by

$I=\large\frac{P_{0}}{4 \pi r^2}=\large\frac{1 W}{4 \pi \;(0.5 m)^2}=0.32 W/m^2$

The target area A is $\;\pi\;(1.3\times10^{-10}m)^2\quad \;5.3\times10^{-20} m $

So the rate at which energy falls on target is given by $\;P=IA=(0.32)\;5.3\times10^{-20}$

$=1.7\times10^{-20}\;J/s$

If all this incoming energy is absorbed , the time required to accumulate enough energy for the electron to escape is

$t=(\large\frac{1.8eV}{1.7\times10^{-20} J/s})\;(\large\frac{1.6\times10^{-19} J}{1eV})=17s\;.$

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