Emission transitions in the Paschen series end at orbit $n=3$ and start from orbit $n$ and can be represented as $\nu = 3.29 \times 10^{15} \, Hz \left(\frac{1}{3^2} - \frac{1}{n^2}\right)$. Calculate the value of $n$ if the transition is observed at $1285 \, nm$. Find the region of the spectrum.

(A) Given wavelength $\lambda = 1285 \, nm = 1285 \times 10^{-9} \, m$.
Frequency $\nu = \frac{c}{\lambda} = \frac{3.0 \times 10^8 \, m/s}{1285 \times 10^{-9} \, m} \approx 2.334 \times 10^{14} \, Hz$.
Using the formula $\nu = 3.29 \times 10^{15} \left(\frac{1}{9} - \frac{1}{n^2}\right)$:
$2.334 \times 10^{14} = 3.29 \times 10^{15} \left(\frac{1}{9} - \frac{1}{n^2}\right)$.
$\frac{2.334 \times 10^{14}}{3.29 \times 10^{15}} = \frac{1}{9} - \frac{1}{n^2}$.
$0.07094 = 0.1111 - \frac{1}{n^2}$.
$\frac{1}{n^2} = 0.1111 - 0.07094 = 0.04016$.
$n^2 = \frac{1}{0.04016} \approx 24.9$.
$n = \sqrt{24.9} \approx 5$.
Since the Paschen series corresponds to transitions ending at $n=3$,and the wavelength $1285 \, nm$ is in the infrared range,the transition is from $n=5$ to $n=3$ and lies in the infrared region.