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If the shortest wavelength of H-atom in Lyman series is `x`, then longest wavelnght in Balmer series of `HE^(2+)` is :A. ` (9x)/5`B. `(36x)/5`C. `x/4`D. `(5x)/9` |
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Answer» Correct Answer - A `1/(lambda _L) = RH [1/1^2 - 1/(prop)]` for shortest ` lambda` of lyman series of H `1/(lambda_B) = Z^2 R_H [1/2^2 -1/3^2]` for longest `lambda` of Bolmer series of ` He^+` ` :. (lambda_B )/(lambda_L) = (1 xx 36)/( 5 xx 4) :. Lambda _B = 9/5 . lambda_L = 9/5 x`. |
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