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The work function of potassium is 2.2 eV. UV light of wavelength 3000 Å and intensity 2 Wm^(-2) is incident on the potassium surface. (i) Determine the maximum kinetic energy of the photo electrons (ii) If 40% of incident photons produce photo electrons, how many electrons are emitted per second if the area of the potassium surface is 2 cm^(2) ? |
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Answer» Solution :(i) The energy of the photon is `E = (hc)/(lambda) = (6.626 xx 10^(-34) xx 3 xx 10^(8))/(3000 xx 10^(-10))` `E = 6.626 xx 10^(-19)`J = 4.14 eV br> Maximum KE of the photoelectrons is `K_(MAX) = h upsilon - phi_(0) = 4.14 - 2.2 = 1.94` eV (ii) The number of photons reaching the surface PER second is `n_(P) = (P)/(E) xx A` `= (2)/(6.626 xx 10^(-19)) xx 2 xx 10^(-4) = 6.04 xx 10^(14)` photons/sec The rate of emission of photoelectrons is `= (0.40) n_(P) = 0.4 xx 6.04 xx 10^(14)` `= 2.415 xx 10^(14)` photoelectrons/sec |
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