The wavelength of the first line in the balmer series is `656 nm `. Calculate the wavelength of the second line and the limiting line in the Balmer series.

Explanation

Wavelength of the second line
According to Rydberg formula :`(1)/(lambda)=bar(v)=R_(H)((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))`
For the first line , `n_(1)=2, n_(2)=3, lambda=656 nm `.
`(1)/(656)= R_(H) ((1)/(2^(2))-(1)/(3^(2)))=R_(H) ((1)/(4)-(1)/(9))=R_(H) xx (5)/(36)=(5R_(H))/(36)`
For the second line , `" " n_(1)=2, n_(2)=4`.
`(1)/(lambda)=R_(H)((1)/(2^(2))-(1)/(4^(2)))=R_(H)((1)/(4)-(1)/(16))=R_(H) xx (3)/(16)=(3R_(H))/(16)`
For the second line ,
Dividing (i) by (ii), `" " (lambda)/(656)=(5)/(36) xx (16)/(3) "or" lambda =(656 xx 5 xx16)/(36 xx 3)=485.9 nm`
Step II. Wavelength of the limiting line
`n_(1)=2, n_(2)=oo , (1)/(lambda)=R_(H)((1)/(2^(2))-(1)/(oo^(2)))=(R_(H))/(4)`
Deviding (i) by (ii) , `" " (lambda)/(656)=(5)/(36)xx4 , lambda =(5 xx4 xx 656)/(36)=364.4 nm`