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The frequency corresponding to transition `n = 1` to `n = 2` in hydrogen atom is.A. `15.66 xx 10^10 Hz`B. `24.66 xx 10^14 Hz`C. `30.57 xx 10^14 Hz`D. `40.57 xx 10^24 Hz` |
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Answer» Correct Answer - B (b) `v = (1)/(lamda) R[(1)/(n_1^2)-(1)/(n_2^2)]` =`109678[(1)/(1) -(1)/(4)] = 82258.5` `lamda = 1.21567 xx 10^-5 cm` or `lamda = 12.1567 xx 10^-6 cm` =`12.1567 xx 10^-8 m` `v = (c)/(lamda) = (3 xx 10^8)/(12.567 xx 10^-8) = 24.66 xx 10^14 Hz`. |
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