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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
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A small potassium foil is placed (perpendicular to direction of incidence of light ) a distance r (=0.5 m) from a point light source whose output power $\;P_{0}\;$ is 1W . Assuming wave nature of light . how long would it take for foil to soak up enough energy (=1.8eV) from the beam to eject on electron ? (Assume that ejected photoelectron collected its energy from a circular area of foil whose radius equals radius of potassium atom ($\;1.3\times10^{-10}m$))

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

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1 Answer

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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
$t=(\large\frac{1.8eV}{1.7\times10^{-20} J/s})\;(\large\frac{1.6\times10^{-19} J}{1eV})=17s\;.$
answered Feb 22, 2014 by yamini.v
 

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