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. `(4)/(9)`
C. `(9)/(4)`
D. `(27)/(5)`

Answer

A

Explanation

For Lyman series
`((1)/(lambda_(max)))_(L) = R(1)^(2) [(1)/((1)^(2)) - (1)/((2)^(2))]`
`(lambda_(max))_(L) = (4)/(3R)`
For Balmer series `((1)/(lambda_(max)))_(B) = R(1)^(2) [(1)/((1)^(2)) - (1)/((3)^(2))]`
`(lambda_(max))_(B) = (36)/(5R)`
`((lambda_(max))_(L))/((lambda_(max))_(B)) = (4)/(3R) xx (5R)/(36) = (5)/(27)`