Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented

Solution:

Wavelength of transition $=1285 \mathrm{~nm}$

$=1285 \times 10^{-9} \mathrm{~m}$ (Given)

$v=3.29 \times 10^{15}\left(\frac{1}{3^{2}}-\frac{1}{n^{2}}\right)$ (Given)

Since $v=\frac{c}{\lambda}$

$=\frac{3.0 \times 10^{8} \mathrm{~ms}^{-1}}{1285 \times 10^{-9} \mathrm{~m}}$

$v=2.33 \times 10^{14} \mathrm{~s}^{-1}$

Substituting the value of $v$ in the given expression,

$3.29 \times 10^{15}\left(\frac{1}{9}-\frac{1}{n^{2}}\right)=2.33 \times 10^{14}$

$\frac{1}{9}-\frac{1}{n^{2}}=\frac{2.33 \times 10^{14}}{3.29 \times 10^{15}}$

$\frac{1}{9}-0.7082 \times 10^{-1}=\frac{1}{n^{2}}$

$\Rightarrow \frac{1}{n^{2}}=1.1 \times 10^{-1}-0.7082 \times 10^{-1}$

$\frac{1}{n^{2}}=4.029 \times 10^{-2}$

$n=\sqrt{\frac{1}{4.029 \times 10^{-2}}}$

$n=4.98$

$n \approx 5$

Hence, for the transition to be observed at $1285 \mathrm{~nm}, n=5$.

The spectrum lies in the infra-red region.