Explanation

(a)
Wave number of spectral line in emission spectrum of hydrogen,
$\overline{v} = R_{H}\left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
Given, $\overline{v} = \frac{8}{9}R_{H}$
On putting the value of $\overline{v}$ in Eq. (i), we get
$\frac{8}{9} = R_{H}\left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)$
$\frac{8}{9} = \frac{1}{(1)^{2}} - \frac{1}{n_{2}^{2}}$
$\frac{8}{9} - 1 = - \frac{1}{n_{2}^{2}}$
$\frac{1}{3} = \frac{1}{n_{2}}\$
$\therefore\ \ \ \ \ n_{2} = 3$
Hence, electron jumps from $n_{2} = 3\ to\ n_{1} = 1$