Question

Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29 × ${10}^{1}5$ (Hz) [1/$3^2$ – 1/$n^2$ ] Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.

Answer 1

Wavelength of transition = 1285 nm
= 1285 × ${10}^{–9}$ m (Given)
$v=3.29\times {10}^{15}(\frac{1}{3^2}-\frac{1}{n^2})since,v=\frac{c}{\lambda }$
=$\frac{3.0\times {10}^{8}m{s}^{-1}}{1285\times {10}^{-9}m}Now,v=2.33\times {10}^{14}{s}^{-1}$
Substituting the value of ν in the given expression,
$3.29\times {10}^{15}(\frac{1}{9}-\frac{1}{n^2})=2.33\times {10}^{14}$
$\frac{1}{9}-\frac{1}{n^2}=\frac{2.33\times {10}^{14}}{3.29\times {10}^{15}}$
$\frac{1}{9}-0.7082\times {10}^{-1}=\frac{1}{n^2}$
$\to \frac{1}{n^2}=1.1\times {10}^{-1}-0.7082\times {10}^{-1}$
$\frac{1}{n^2}=4.029\times {10}^{-2}$
$n=\sqrt{\frac{1}{4.029\times {10}^{-2}}}n=4.98n\approx 5$
Hence, for the transition to be observed at 1285 nm, n = 5.
The spectrum lies in the infra-red region.