Radius of \(n\)th orbit of \(\mathrm{H}\)-like particles
\(\begin{aligned}
&=\frac{0.529 \mathrm{n}^2}{Z} Å=\frac{52.9 \mathrm{n}^2}{Z} \mathrm{pm} \\
&r_1=1.3225 \mathrm{~nm}=1322.5 \mathrm{pm}=52.9 \mathrm{n}_1^2 \\
&r_2=211.6 \mathrm{pm}=\frac{52.9 \mathrm{n}_2^2}{Z} \\
&\therefore \frac{r_1}{r_2}=\frac{1322.5}{211.6}=\frac{n_1^2}{n_2^2}=6.25 \text { or } \frac{n_1}{n_2}=2.5
\end{aligned}\)
\(\therefore\) If \(\mathrm{n}_2=2, \mathrm{n}_1=5\). Thus, the transition is from 5 th orbit. It belongs to Balmer series.
\(\bar{v}=1.097 \times 10^7 \mathrm{~m}^{-1}\left(\frac{1}{2^2}-\frac{1}{5^2}\right)\) \(=1.097 \times \frac{21}{100} \times 10^7 \mathrm{~m}^{-1}\) or \(\lambda=\frac{1}{\bar{v}}=\frac{100}{1.097 \times 21 \times 10^7} \mathrm{~m}\) \(=434 \times 10^{-9} \mathrm{~m}=434 \mathrm{~nm}\) Thus, it lies in the visible region