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Calculate the kinetic energy of the electron ejected when yellow light of frequency `5.2 xx 10^(14) sec^(-1)` falls on the surface of potassium metal. Threshold frequency of potassium is `5 xx 10^(14) sec^(-1)` |
Answer» K.E. of the ejected electron is given by `(1)/(2) mv^(2) = hv - hv_(0) = h (v - v_(0))` `= 6.625 xx 10^(-34) Js (5.2 xx 10^(14) - 5.0 xx 10^(14)) s^(-1) = 6.625 xx 10^(-34) xx 0.2 xx 10^(14)` joules `= 1.325 xx 10^(-20)` Joules |
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