In the spectrum of hydrogen atom, the ratio of the longest wavelength in Lyman series to the longest wavelangth in the Balmer series is:A. `(5)/(27)`B. `(4)/(9)`C. `(9)/(4)`D. `(27)/(5)`

In the spectrum of hydrogen atom, the ratio of the longest wavelength

1.	In the spectrum of hydrogen atom, the ratio of the longest wavelength in Lyman series to the longest wavelangth in the Balmer series is:A. `(5)/(27)`B. `(4)/(9)`C. `(9)/(4)`D. `(27)/(5)`
Answer» Correct Answer - A For Lyman series `((1)/(lambda_(max)))_(L) = R(1)^(2) [(1)/((1)^(2)) - (1)/((2)^(2))]` `(lambda_(max))_(L) = (4)/(3R)` For Balmer series `((1)/(lambda_(max)))_(B) = R(1)^(2) [(1)/((1)^(2)) - (1)/((3)^(2))]` `(lambda_(max))_(B) = (36)/(5R)` `((lambda_(max))_(L))/((lambda_(max))_(B)) = (4)/(3R) xx (5R)/(36) = (5)/(27)`

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In the spectrum of hydrogen atom, the ratio of the longest wavelength in Lyman series to the longest wavelangth in the Balmer series is:A. `(5)/(27)`B. `(4)/(9)`C. `(9)/(4)`D. `(27)/(5)`

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