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Light of wavelength 5000Å falls on a sensitive plate with photoelectric work function of 1.9 eV. The K.E. of the photo electron emitted will be |
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Answer» Solution :Here `lambda = 5000 Å = 5 xx 10^(-7)m` `W_(0) = 1.9 EV` (i) Energy of a photon, `E = (hc)/(lambda) = (6.6 xx 10^(-34) xx 3 xx 10^(8))/(5 xx 10^(-7))J = (6.6 xx 10^(-34) xx 3 xx 10^(8))/(5 xx 10^(-7) xx 1.6 xx 10^(-19))eV` E = 2.475 eV (ii) K.E of a photoelectron, `K.E = h UPSILON - W_(0) = 2.475 - 1.9 = 0.575 eV` (iii) LET `V_(0)` be the STOPPING potential. Then `eV_(0) = (1)/(2) mv^(2) =` K.E of a photoelectron `V_(0) = (0.575)/(e) eV` `V_(0) = 0.575 V` |
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