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The time period of revolution of an electron in its ground state orbit in a hydrogen atom is 1.6xx10^(-16) s. The frequency of the electron in its first excited state (in s^(-1) ) is : |
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Answer» `7.8xx10^(14)` Suppose, time period is `T_(1)` in ground state and `T_(2)` in first excited state. `:.n_(1)=1` and `T_(1)=1.6xx10^(-16)s` and `n_(2)=2` and `T_(2)=?` `:(T_(2))/(T_(1))=((n_(2))/(n_(1)))^(3)""` (z=1 same ) `:.T_(2)=T_(1)XX((2)/(1))^(3)` `=1.6xx10^(16)xx8` `=12.8xx10^(-16)` `:.` FREQUENCY `v_(2)=(1)/(T_(2))=(1)/(12.8xx10^(-16))` `:.v_(2)=0.0781xx10^(16)` `:.v_(2)~~7.8xx10^(14)s^(-1)` |
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