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 × 1015 (Hz) [$\frac{1}{3^2}-\frac{1}{n^2}$] 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 × 109m (Given)
v = 3.29 × 1015 ($\frac{1}{3^2}-\frac{1}{n^2}$) (Given)
Since v = c / λ = $\frac{3.0 × 10^8 ms ^{- 1} }{1285 × 10^{-9} m}$
v = 2.33 × 1014 s-1
Substituting the value of v in the given expression,
3.29 × 1015 ($\frac{1}{9}-\frac{1}{n^2}$) = 2.33 × 1014
$\frac{1}{9}-\frac{1}{n^2}$=$\frac{2.33\times10^{14}}{3.29\times 10^{15}}$
$\frac{1}{9}$ - 0.7082 × 10 - 1 = $\frac{1}{n^2}$
⇒ $\frac{1}{n^2}$ = 1 .1 × 10 -1 - 0.7082 × 10 -1
$\frac{1}{n^2}$ = 4.029 × 10 - 2
n = $\sqrt{\frac{1}{4.029}}\times10^{-2}$
n= 4.98 n
Hence, for the transition to be observed at 1285 nm, n= 5.
The spectrum lies in the infra-red region.