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The frequency of the incident light falling on a photosensitive metal plate is doubled. The kinetic energy of the emitted photoelectron is : |
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Answer» unchanged `:. (E_(k).)/(E_(k))=(2hv-phi_(0))/(hv-phi_(0))=(2((hv-phi_(0))/(2)))/(hv-phi_(0))s` Clearly `(E_(k).)/(E_(k)) GT 2` so `E_(k). gt 2E_(k)` |
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