calculate the longest wavelength in Balmer series and the series limit . (Given `R=1.097xx10^(7)m^(-1)` )

Explanation

`overline(v)=(1)/(lamda)=R((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))`
For shortest wavelength `n_(1)=2,n_(2)=oo`
Substitution with final answer `lamda=3646Å or lamda=3.646xx10^(-7)m`
For longest wavelength `n_(1)=2,n_(2)=3`
Substitution with final answer `lamda=6563Å or lamda=6.563xx10^(-7)m`
Important Note:
Any other relevant or alternate answer may be considered.
Detailed Answer:
The lines of Balmer series is given by
`(1)/(lamda)=R[(1)/(2^(2))-(1)/(n_(2)^(2))]`
For short wavelength, `n=-oo`
`(1)/(lamda)=R[(1)/(4)-(1)/((-oo)^(2))]`
`(1)/(lamda)=(R)/(4)`
`Rlamda=4`
`lamda=(4)/(1.097xx10^(7))`
`lamda=(4)/(1.097)xx10^(-7)`
`lamda=3646Å`,
For long wavelength
`(1)/(lamda_("long"))=R[(1)/((2)^(2))-(1)/((3)^(2))]`
`(1)/(lamda_("long"))=R[(1)/(4)-(1)/(9)]`
`(1)/(lamda_("long"))=R[(9-4)/(36)]`.
`(1)/(lamda_("long")=R[(5)/(36)]`
`lamda(5R)=36`
`lamda=(36)/(5R)xx10^(-7)=(36)/(5xx1.097xx10^(+7))`
`lamda_("long")=6.563xx10^(-7)=6563`Å