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In a photoelectric set up, a point source of light of power 3.2xx10^(-3) W emits monoenergetic photons of energy 5.0 eV. The source is located at a distance of 0.8 m from the centre of a stationary metallic sphere of work function 3.0 eV and of radius 8.0xx10^(-3) m. The efficiency of photoelectron emission is one for every 10^(6) incident photon. Assume that state is isolated and initially neutral and that the photoelectrons are instantly swept away after emission. Calculate the number of photoelectron emitted per sec, by sphere : |
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Answer» `10^(2)` `P=2.2xx10^(-3)W` Energy of onephoton `E=5.0eV=5xx1.6xx10^(-19)J=8.0xx10^(-19)J` Number of photon emitted PER sec by source, `N=(P)/(E)=(3.2xx10^(-3))/(8.0xx10^(-19))=4xx10^(15)` If `r=0.8 m` is the DISTANCE of the sphere from the source, then the number of photons incident on the sphere per sec unit area. `=(N)/(4pir^(2))` If `R=8.0xx10^(-3)m` is the radius of the sphere, the number of photons incident on sphere per sec. `=(N)/(4pir^(2))xx4pi R^(2)=(NR^(2))/(4pir^(2))xx4` `=(4xx10^(15)xx(8.0xx10^(-3))2)/(4XX(0.80)^(2))=4xx10^(11)` As efficiency of photoelectron EMISSION is 1 for every `106(6)` incident photons, therefore, the number of emitted photoelectrons per sec `=4xx(10^(11))/(10^(6))=4xx10^(5)~~10^(5)` |
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