To what series does the spectral lines of atomic hydrogen belong if its wavenumber is equal to the difference between the wavenumbers of the following two lines of the Balmer series: `486.1` and `419.2 nm`? What is the wavelength of that line?

Answer

`n_(1)-4,n_(2)=6,2.63xx10^(-4)cm`

Explanation

Given that `lambda_(1)=486.1xx10^(-9)m`
`=486.1xx10^(-7)cm`
`lambda_(2)=410.2xx10^(-9)m=410.2xx10^(-7)cm`
and `bar(v)=bar(v)_(2)-bar(v)_(1)=[(1)/(lambda_(2))-(1)/(lambda_(1))]`
`R_(H)=[(1)/(2^(2))-(1)/n_(2)^(2)]-R_(H)[(1)/(2^(2))-(1)/(n_(1)^(2))]`
`v=R_(H)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]" "....(i)`
For line 1 of Balmer series
`(1)/(lambda_(1))=R_(H)[(1)/(2^(2))-(1)/(n_(1)^(2))]=109678[(1)/(2^(2))-(1)/(n_(1)^(2))]`
or `(1)/(456.1xx10^(-7))=109678[(1)/(2^(2))-(1)/(n_(1)^(2))]`
`therefore n_(1)=4`
For line II of Balmer series ,
`(1)/(lambda_(1))=R_(H)[(1)/(2^(2))-(1)/(n_(2)^(2))]=109678[(1)/(2^(2))-(1)/(n_(2)^(2))]`
or `(1)/(410.2xx10^(-7))=109678[(1)/(2^(2))-(1)/(n_(2)^2)]`
`therefore n_(2) =6`
Thus given electronic transition occurs from `6^(th)` to `4^(th)` shell. Also by eq.(i)
`bar(v)=(1)/(lambda)=109678[(1)/(4^(2))-(1)/(6^(2))]`
`therefore lambda=2.63xx10^(-4)cm`