Explanation

(c) The wavelength of different members of Balmer series are given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{\lambda}$ $= R_{H}\left\lbrack \frac{1}{2^{2}} - \frac{1}{n_{i}^{2}} \right\rbrack$, where $n_{i}$= 3,4,5,….
The first member of Balmer series ($H_{\alpha})$ corresponds to $n_{i}$=3.It has maximum energy and hence the longest wavelength. Therefore ,wavelength of $H_{\alpha}$ line (or longest wavelength )
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{\lambda_{1}}$ $= R_{H}\left\lbrack \frac{1}{2^{2}} - \frac{1}{3^{2}} \right\rbrack$
$\ \ \ \ \ \ \ \ \ \ \ \ = 1.097 \times \ 10^{7}\left( \frac{5}{36} \right)$
or$\ \ \ \ \ \ \ \ \lambda_{1}$= $\frac{36}{5\ \times \ 1.097\ \times 10^{7}}$ $= \ 6.563\ \times \ 10^{- 7}\ m$
$\ \ \ \ \ \ \ \ \ \ \ \ n = \ 6563\ \mathring{\mathrm{A}}$
The wavelength of the Balmer series limit corresponds to $n_{i}\ = \ \infty$ and has got shortest wavelength.
Therefore , wavelength of Balmer series limit is given by
$\ \ \ \ \ \ \ \ \ \frac{1}{\lambda_{\infty}}$ =$R_{H}\left\lbrack \frac{1}{2^{2}} - \frac{1}{\infty^{2}} \right\rbrack$ $= 1.097\ \times \ 10^{7}\ \times \ \frac{1}{4}$
or$\ \ \ \lambda_{\infty} = \frac{4}{1.097\ \times 10^{7}}$ =$3.646\ \times \ 10^{- 7}\ m$
$= 3646\mathring{\mathrm{A}}$
Only $4861\ \mathring{\mathrm{A}}\ $is between the first and last line of the Balmer series.