What is the series limit for the Balmer series? (The series limit, like the name...

Rydberg is our knight in shining armour, again! We know, for the Balmer series $n_1=2$. The series limit, or the very last line in the series, will be obtained by putting $n_2=\mathrm{\infty }$. Thus the Balmer series limit can be found as –
$\frac{1}{{\lambda }_{\mathrm{\infty }}}=R(\frac{1}{2^2}-\frac{1}{{\mathrm{\infty }}^2})=\frac{R}{4}{m}^{-1}$
}\Rightarrow \lambda_\infty=\frac4Rm=3.646\times10^{−7}m=364.6nm.Show External URL Show Embeded Code Hide MathML Code⇒λ∞=4Rm=3.646×10−7m=364.6nm.$\Rightarrow {\lambda }_{\mathrm{\infty }}=\frac{4}{R}m=3.646\times {10}^{-7}m=364.6nm.$
Option (a) is correct!

The significance of the wavelength is, it’s the Balmer line with the smallest wavelength. The next line with a smaller wavelength will be the first line of the Lyman series, which is 121.6 nm. This essentially means, there will be no spectral lines in the range 121.6 nm - 364.6nm.