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When a metal is illuminated with light of frequency f, the maximum kinetic energy of the photoelectrons is 1.2 eV. When the frequency is increased by 50% the maximum kinetic energy increases to 4.2 eV. What is the threshold frequency for this metal? |
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Answer» Correct Answer - A `K_(max) = E-W 1.2 = E-W …(i) 4.2 = 1.5 E - W …..(ii) Solving these equations, we get `w = 4.8 eV = hf_0` `:. f_0 = (4.8xx1.6xx10^(-19)/6.63xx10^(-34)` `=1.16xx10^15 Hz.` |
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