Calculate the wavelength for the emission transition if it starts from the orbit having radius $1.3225 \, nm$ and ends at $211.6 \, pm$. Name the series to which this transition belongs and the region of the spectrum.

(D) The radius of the $n^{\text{th}}$ orbit of a hydrogen-like species is given by $r = 0.0529 \times \frac{n^2}{Z} \, nm = 52.9 \times \frac{n^2}{Z} \, pm$.
For the initial orbit $(r_1 = 1.3225 \, nm = 1322.5 \, pm)$:
$n_1^2 = \frac{1322.5 \times Z}{52.9} = 25Z$.
For the final orbit $(r_2 = 211.6 \, pm)$:
$n_2^2 = \frac{211.6 \times Z}{52.9} = 4Z$.
Taking the ratio: $\frac{n_1^2}{n_2^2} = \frac{25Z}{4Z} = 6.25$.
$\frac{n_1}{n_2} = \sqrt{6.25} = 2.5 = \frac{5}{2}$.
Thus, $n_1 = 5$ and $n_2 = 2$.
The transition is from $n = 5$ to $n = 2$, which belongs to the $\text{Balmer}$ series, and the spectral region is the visible region.
Using the Rydberg formula for wave number $(\bar{\nu})$:
$\bar{\nu} = R_H \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) = 1.097 \times 10^7 \, m^{-1} \left( \frac{1}{2^2} - \frac{1}{5^2} \right)$.
$\bar{\nu} = 1.097 \times 10^7 \times \left( \frac{1}{4} - \frac{1}{25} \right) = 1.097 \times 10^7 \times \frac{21}{100} = 2.3037 \times 10^6 \, m^{-1}$.
Wavelength $(\lambda)$ = $\frac{1}{\bar{\nu}} = \frac{1}{2.3037 \times 10^6} \, m \approx 4.34 \times 10^{-7} \, m = 434 \, nm$.