The wavelength of `H_(beta)` spectral line of Balmer series is 4860 Å then find out the wavelength of `H_(alpha)` spectral line of the same series

Explanation

For `H_(beta)` line of Balmer series of hydrogen atom H
`(1)/(lambda_(beta))=R[(1)/(2^(2))-(1)/(4^(2))]=R[(1)/(4)-(1)/(16)]`
`(1)/(lambda_(beta))=(3R)/(16).....(1)`
`rArr` For `H_(alpha)` line of Balmer series of hydrogen atom
`(1)/(lambda_(alpha))=R[(1)/(2^(2))-(1)/(3^(2))]=R[(1)/(4)-(1)/(9)]`
`(1)/(alpha)=(5R)/(36)...(2)`
`rArr` Taking the ratio of equation (1) and (2)
`(lambda_(alpha))/(lambda_(beta))=(3R)/(16)xx(36)/(5R)`
`lambda_(alpha)=lambda_(beta)xx(27)/(20)`
`=4860xx(27)/(20)`
` :. lambda_(alpha)=6561Å`