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The threshold frequency for certain metal is `v_0`. When light of frequency `2v_0` is incident on it, the maximum velocity of photoelectrons is `4xx10^(6) ms^(-1)`. If the frequency of incident radiation is increaed to `5v_0`, then the maximum velocity of photoelectrons will beA. `4xx10^(6) m//sec`B. `6xx10^(6) m//sec`C. `8xx10^(6) m//sec`D. `16xx10^(6) m//sec` |
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Answer» Correct Answer - C `h.2 upsilon_(0) = h upsilon_(0) +(1)/(2)m(4xx10^(6))^(2)` ….(i) `h.5 upsilon_(0) = h upsilon_(0) +(1)/(2)m v_(max)^(2)` …..(ii) `(v_(max)^(2))/((4xx10^(6))^(2)) =4 implies v_(max) =8xx10^(6) m//sec` |
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