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The longest ` lambda` for the Lyamn series is …………… (Given ` R_H = 109678 cm^(-1))` :A. ` 1215`B. ` 1315`C. `1415`D. `1515` |
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Answer» Correct Answer - A For longest of Lyman series . ` n_(2) = 2`, `(1)/( lambda) = R_(H) [(1)/(n_(1)^(2)) - (1)/n_(2)^(2)]` because ` Delta E = (hc)/( lambda)` is minimum when `lambda` is longest , Thus ` Delta E = E_(2) - E_(1)` = minimum. |
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