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A uniform monochromatic beam of light of wavelength 365xx10^-9m and intensity 10^-8 W m^-2 falls on a surface having absorption coefficient 0.8 and work function 1.6 eV. Determine the number of electrons emitted per square metre per second, power absorbed per m^2, and the maximum kinetic energy of emitted photo electrons. |
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Answer» Solution :Let N be the number of PHOTOELECTRON per second. `"Intensity"=Nxx"energy of one photon"` `I=N(hc)/(lamda)` Number of incident photons, `N_i="insident i ntensity"XX(lamda)/(hc)` `=(10^(-8)xx365xx10^(-9))/(6.62xx10^(-34)xx3xx10^(8))=18.35xx10^(8)` Number of photon `N_(ab)` absorbed by the surface per unit area per unit time (i.e., absorbed FLUX), `N_(ab)=0.8xx18.35xx10^(9)` `=1.47xx10^(10)m^(-2)s^(-1)` Emitted flux `=` No. of electrons emitted `m^(-2)s^(-1)` `=1.47xx10^(10)m^(-2)s^(-1)` Power absorbed per `m^(2)=0.8xx` Incident intensity `=0.8xx10^(-8)=8xx10^(-9)Wm^(-2)` From Einstein's PHOTOELECTRIC equation, `(KE)_(max)=hv-W_(0)=(hc)/(lamda)-W_(0)` `(6.62xx10^(-34)xx3xx10^(8))/(365xx10^(-9))-1.6xx1.6xx10^(-19)` `=2.89xx10^(-19)J=1.80eV` |
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