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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. `(4)/(5)xx10^(6)ms^(-1)`B. `2xx10^(6)ms^(-1)`C. `8xx10^(6)ms^(-1)`D. `2xx10^(7)ms^(-1)` |
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Answer» Correct Answer - C In the first case, `(1)/(2)mv_(max)^(2)=2hupsilon_0-hupsilon_0=hupsilon_0` In the second case, `(1)/(2)mv_(max)^(2)=5hupsilon_0-hupsilon_0=4hupsilon_0` Clearly, `v_(max)` is doubled. |
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