The wave number of the spectral line in the emission spectrum of hydrogen will be equal to \(\frac{8}{9}\) times the Rydberg's constant if the electron jumps from

Wave number of spectral line in emission spectrum of hydrogen,
\[
\bar{v}=R_{H}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \quad \text{...(i)}
\]
Given, \(\quad \bar{v}=\frac{8}{9} R_{H}\)
On putting the value of \(\bar{v}\) in Eq. (i), we get
\[
\begin{aligned}
\frac{8}{9} \mathrm{R}_{\mathrm{H}} &=\mathrm{R}_{\mathrm{H}}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right) \\
\frac{8}{9} &=\frac{1}{(1)^{2}}-\frac{1}{\mathrm{n}_{2}^{2}} \\
\frac{8}{9}-1 &=-\frac{1}{\mathrm{n}_{2}^{2}} \\
\frac{1}{3} &=\frac{1}{\mathrm{n}_{2}} \\
\therefore \quad \mathrm{n}_{2} &=3
\end{aligned}
\]
Hence, electron jumps from \(\mathrm{n}_{2}=3\) to \(\mathrm{n}_{1}=1\)