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Light of wavelength `5000 Å` falls on a sensitive plate with photoelectric work function of `1.9 eV`. The kinetic energy of the photoelectron emitted will beA. `0.58 eV`B. `2.48 eV`C. `1.24 eV`D. `1.16 eV` |
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Answer» Correct Answer - A `E = W_(0) + K_(max) , E = (12375)/(5000) = 2.475 eV` `:. K_(max) = E - W_(0) = 2.475 - 1.9 = 0.57 eV` |
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