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Q.

A small potassium foil is placed (perpendicular to the direction of incidence of light) a distance r (= 0.5 m) from a point light source whose output power P0 is 1.0 W. Assuming wave nature of light how long (in s) would it take for the foil to soak up enough energy (= 1.8 eV) from the beam to eject an electron? Assume that the ejected photoelectron collected its energy from a circular area of the foil whose radius equals the radius of a potassium atom (1.3 x 10-10 m).

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answer is 17.

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Detailed Solution

If the source radiates uniformly in all directions, the intensity I of the light at a distance r is given byI=P04πr2=1.0W4π(0.5m)2=0.32W/m2The target area A is n(1.3 x 10-10 m)2 or 5.3 x 10-20m2, so that the rate at which energy falls on the target is given byP=IA=0.32W/m25.3×10−20m2=1.7×10−20J/sIf all this incoming energy is absorbed, the time required to accumulate energy for the electron to escape ist=1.8eV1.7×10−20J/s1.6×10−191eV=17s
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