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

Options

A. `(4)/(3)`
B. `(36)/(5)`
C. `(1)/(4)`
D. `(5)/(9)`

Answer

A

Explanation

(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)`.