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In a photoelectric experiment, with light of wavelength `lambda`, the fastest election has speed v. If the exciting wavelength is changed to `(3lambda)/4`, the speed of the fastest emitted electron will becomeA. `= v((4)/(3))^(1/2)`B. `= v((3)/(4))^(1/2)`C. `gt v((4)/(3))^(1/2)`D. `lt v((4)/(3))^(1/2)` |
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Answer» Correct Answer - C `hv_(0)^(2) - hv_(0) = (1)/(2) m v^(2)` `:. (4)/(3) hv_(0) - hv_(0) = (1)/(2) m v^(-2)` `:. (v^(2))/(v^(2) = ((4)/(3) v - v_(0)))/(v - v_(0)) :. V ^(1) = v sqer(((4)/(3) v - v_(0)))/(v - v_(0)))` `:. V^(1) gt sqrt((4)/(3))` |
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