The extreme wavelengths of Paschen series are
Answer & explanation
Correct answer: option 2
ln Paschen series $\frac{1}{\lambda_{\max }}=R\left[\frac{1}{(3)^2}-\frac{1}{(4)^2}\right]$
$\Rightarrow \lambda_{\max }=\frac{144}{7 R}=\frac{144}{7 \times 1.1 \times 10^7}=1.89 \times 10^{-6} \mathrm{~m}=1.89 \mu \mathrm{m}$
Similarly $\lambda_{\min }=\frac{9}{R}=\frac{9}{1.1 \times 10^7}$ = 0.818 µm