The wavelength of the first line of Lyman series is `lambda` for hydrogen atom. Find out the wavelength of first line of Pfund series.

Explanation

If `lambda` is the wavelength of first line in Lyman series then
`(1)/(lambda)=R[(1)/(1^(2))-(1)/(2^(2))]=R[(1)/(1)-(1)/(4)]`
`(1)/(lambda)=(3R)/(4) ....(1)`
`rArr` is the wavelength of first line in Pfund series then,
`(1)/(lambda.)=R[(1)/(5^(2))-(1)/(6^(2))]=R[(1)/(25)-(1)/(36)]`
`(1)/(lambda.)= (11F)/(900)...(2)`
`rArr` Taking ratio of equation (1) and (2)
`(lambda.)/(lambda)=(3R)/(4)xx(900)/(11R)=(2700)/(44)`
`lambda.=61.36 lambda`