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Light of intensity 10^(-5) W m^(-2) falls on a sodium photo-cell of surface area 2 cm^(2) . Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer? |
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Answer» Solution :ASSUME one conduction electron per atom. Effective atomic area `~10^(-20)m^(2)` Number of electrons in 5 layers `=(5xx2xx10^(-4)m^(2))/(10^(-20)m^(2))=10^(17)` Incident power `=10^(-5)Wm^(-2)xx2xx10^(-4)m^(2)=2xx10^(-9)W` In the wave picture, incident power is uniformly absorbed by all the electrons continuously. Consequently, energy absorbed per second per electron `=2xx10^(-9)//10^(17)=2xx10^(-26)W` Time required for photoelectric emission `=2xx1.6xx10^(-19)J//2xx10^(-26)W=1.6xx10^(7)s` Which is about 0.5 year Implication: Experimentally, photoelectric emission is observed nearly instantaneously `(~10^(-9) s)`: Thus, the wave picture is in gross disagreement with experiment. In the photon-picture, energy of the radiation is not continuously shared by all the electrons in the top layers. RATHER, energy comes in discontinuous ‘quanta’. and ABSORPTION of energy does not take place gradually. A photon is either not absorbed, or absorbed by an electron nearly instantly. |
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