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

Explanation

The wavelength of a spectral line in the Lyman series is
`(1)/(lambda_(L))=R((1)/(1^(2))-(1)/(n^(2))),n=2,3, 4, …………..`
and that in the Balmer series is
`(1)/(lambda_(B))=R((1)/(2^(2))-(1)/(n^(2))),n=3,4, 5,…………`
For the longest wavelength in the Lyman series, `n= 2`
`therefore" "(1)/(lambda_(L))=R((1)/(1^(2))-(1)/(2^(2)))=R((1)/(1)-(1)/(4))=R((4-1)/(4))=(3R)/(4)`
`or lambda_(L)=(4)/(3R)`
For the longest wavelength in the Balmer series, n = 3
`therefore" "(1)/(lambda_(B))=R((1)/(2^(2))-(1)/(3^(2)))=R((1)/(4)-(1)/(9))=R((9-4)/(36))=(5R)/(36)`
`"or "lambda_(B)=(36)/(5R)`
Thus, `(lambda_(L))/(lambda_(B))=((4)/(3R))/((36)/(5R))=(4)/(3R)xx(5R)/(36)=(5)/(27)`