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The shortest wavelength of H-atom in Lyman series is x, then longest wavelength in Balmer series of `He^(+)` isA. `(4)/(3)`B. `(36)/(5)`C. `(1)/(4)`D. `(5)/(9)` |
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Answer» Correct Answer - A (a) Longest wavelength in Lyman Series of hydrogen atom arisis from transition between `n = 2` to `n = 1` whose number is given by `overline v_(H(2rarr1)) = R xx 1^2 ((1)/(1^2) - (1)/(2^2)) = (3)/(4) R` Shortest wavelength in Balmer series of `He^+` arises from transition between `n = a` to `n = 2` whose wave number is given by `overline v_(He(alpha -2)) = R xx 2^2 ((1)/(2^2) -(1)/(alpha^2)) = R` =`(overlinev_(He))/(overlinev_H) = (lamda_H)/(lamda_(He))` `:. (lamda_H)/(lamda_(He)) = (4R)/(3R) = (4)/(3)`. |
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