Explanation

(a) When an atom comes down from some higher energy level to the second energy ($n$=2), then the lines of spectrum are obtained in visible part and give the Balmer series.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{\lambda}$ =$\ R\left( \frac{1}{2^{2}} - \frac{1}{n^{2}} \right),n = 3,4,5,\ldots..$
For second line $n = 4$
$\therefore\ \ \ \ \ \ \ \ \ \frac{1}{\lambda}$ =$\ R\left( \frac{1}{2^{2}} - \frac{1}{4^{2}} \right) = \frac{3R}{16}$
$\lambda = \ \frac{16}{3R}$
$\ \ \ \ \ \ \ \ \ \ R\ =$ 1.097 $\times$ 10$^{7}$ m$^{-1}$
$\ \ \ \ \ \ \ \ \ \ \lambda = \frac{16}{3\ \times 1.097\ \times 10^{7}}$
$= \ 4860\ \times \ 10^{- 10}\ m$
$\Rightarrow \ \ \ \ \ \lambda = 4860\ \mathring{\mathrm{A}}$
which corresponds to colour blue.