The shortest wavelength of H atom in Lyman series is x, then the longest wavelength of `He^(+)` in Balmer series is

Options

A. `(9x)/5`
B. `(36x)/5`
C. `x/4`
D. `(5x)/9`

Answer

A

Explanation

`1/lambda = R_H Z^2 [1/(n_1^2)-1/(n_2^2)]`
For Lyman series, `n_1 = 1,lambda` is the shortest if `n_2 = oo`.
`:. barR_H=1/x " for " H- " atom " (Z = 1)`
For Balmer series, `n_1= 2, lambda` is the longest if `n_2 = 3`.
`1/(lambda_("max"))=(1/x)(2)^2[1/2^2 -1/3^2] " for " He^(+) " ion " (Z = 2)`.
`:. lambda_("max")=(9x)/5`.