Wavelenght of the Blamer `Halpha` line is 6565Å. Calculate the wavelenght of `H_(beta)`, line of same hydrogen like atom.

Explanation

For `alpha `-line of Balmer Series
`(1)/lambda_(alpha)=RZ^(2)[(1)/2^(2)-
(1)/3^(2)]`
`beta `-line of Balmer Series
`(1)/lambda_(beta)=RZ^(2)[(1)/2^(2)-
(1)/4^(2)]`
`lambda_(beta)/(lambda_(alpha))`=`[[(1)/2^(2)-(1)/4^(2)]]/[[(1)/2^(2)-(1)/3^(2)]]`
`lambda_(beta)=4.863xx10^(-7)m`