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When a 6000Å light falls on the cathode of a photo cell and produced photoemission. If a stopping potential of 0.8 V is required to stop emission of electron, then determine the frequency of the light |
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Answer» Solution :Wavelength, `lambda = 6000 Å = 6000 xx 10^(-10) m` stopping potential, `V_(0) = 0.8 V` (i) Frequency of the light, `upsilon = (c)/(lambda)` `= (3 xx 10^(8))/(6000 xx 10^(-10)) = 5 xx 10^(-4) xx 10^(18)` `upsilon = 5 xx 10^(14)` Hz (II) Energy of the INCIDENT photon, `E = h upsilon = 6.6 xx 10^(-34) xx 5 xx 10^(14)` `= 33 xx 10^(-20)J` `= (33 xx 10^(-20))/(1.6 xx 10^(-19)) = 20.625 xx 10^(-1)` E = 2.06 eV (iii) Work function of the cathode material, `W_(0) = h upsilon - e V_(0)` `= ((6.6 xx 10^(-34) xx 5 xx 1-^(14))/(1.6 xx 10^(-19))) - ((1.6 xx 10^(-19) xx 0.8)/(1.6 xx 10^(-19))) = 2.06 - 0.8` `W_(0) = 1.26 eV` (iv) Threshold frequency, `W_(0) = h upsilon_(0)` `upsilon_(0) = (W_(0))/(h) = (1.26 xx 1.6 xx 10^(-19))/(6.6 xx 10^(-34)) = 0.3055 xx 10^(15)` `upsilon_(0) = 3.05 xx 10^(14)` Hz (v) Net energy of the electron after it leaves the surface `E = (upsilon - upsilon_(0))` `= 6.6 xx 10^(-34) (5 xx 10^(14) - 3.06 xx 10^(14))` `= 6.6 xx 10^(-34) xx 1.94 xx 10^(14)` `E = 12.804 xx 10^(-20) J` `= (1.2804 xx 10^(-19))/(1.6 xx 10^(-19))` E = 0.8 eV |
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