In the spectrum of hydrogen atom, the ratio of the longest wavelength in Lyman series to the longest wavelangth in the Balmer series is

Options

A. `(5)/(27)`
B. `(5)/(48)`
C. `(27)/(5)`
D. `(1)/(3)`

Answer

A

Explanation

For Lyman series `(1)/(lamda_(a))= R ((1)/(1^(2))- (1)/(2^(2)))`
For Balmer series - `(1)/(lamda_(a))= R ((1)/(2^(2))- (1)/(3^(2)))`
`((lamda_(alpha))_("Balmer"))/((lamda_(alpha))_("Lyman")) = ((3)/(4))/(5//36) = (3)/(1) xx (9)/(5) = (27)/(5) rArr ((lamda_(alpha))_("Lyman"))/((lamda_(alpha))_("Balmer")) = (5)/(27)`