Question:
What is the shortest wavelength present in the Paschen series of spectral lines?
Solution:
Rydberg’s formula is given as:
$\frac{h c}{\lambda}=21.76 \times 10^{-19}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$
Where,
h = Planck’s constant = 6.6 × 10−34 Js
c = Speed of light = 3 × 108 m/s
(n1 and n2 are integers)
The shortest wavelength present in the Paschen series of the spectral lines is given for values n1 = 3 and n2 = ∞.
$\frac{h c}{\lambda}=21.76 \times 10^{-19}\left[\frac{1}{(3)^{2}}-\frac{1}{(\infty)^{2}}\right]$
$\lambda=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8} \times 9}{21.76 \times 10^{-19}}$
$=8.189 \times 10^{-7} \mathrm{~m}$
$=818.9 \mathrm{~nm}$