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Light of wavelength 390 nm is directed at a metal electrode. To find the energy of electrons ejected, an opposing potential difference is established between it and another electrode. The current of photoelectrons from one to the other is stopped completely when the potential difference is 1.10 V. Determine (i) the work function of the metal and (ii) the maximum wavelength of light that can electrons from this metal. |
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Answer» Solution :(i) The work function is GIVEN by `phi_(0) = h UPSILON - K_(MAX) = (hc)/(lambda) - eV_(0)` SINCE `K_(max) = eV_(0)` `= [(6.626 xx 10^(-34) xx 3 xx 10^(8))/(390 xx 10^(-9))] - [1.6 xx 10^(-19) xx 1.10]` `= 5.10 xx 10^(-19) - 1.76 xx 10^(-19) = 3.34 xx 10^(-19)` J = 2.09 eV (ii) The threshold WAVELENGTH is `lambda_(0) = (hc)/(phi_(0)) = (6.626 xx 10^(-34) xx 3 xx 10^(8))/(3.33 xx 10^(-19))` `= 5.969 xx 10^(-7) m = 5963 Å` |
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