What is the shortest wavelength present in the Paschen series of spectral lines?
Rydberg’s formula is given as: hc/λ = 21.76 x 10-19 [1/(n1)2 - 1/(n2)2]
Where, h = Planck’s constant = 6.6 × 10 −34 Js
c = Speed of light = 3 × 10 8 m/s
(n 1 and n 2 are integers)
The shortest wavelength present in the Paschen series of the spectral lines is given for values n 1 = 3 and n 2 = ∞.
hc/λ = 21.76 x 10-19 [1/(3)2 - 1/(∞)2]
∴ λ = 6.6 x 10-34 x 3 x 108 x 9 / 21.76 x 10-19 = 8.189x10-7 m = 818.9 nm
Hence, the shortest wavelength present is 818.9 nm.
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Welcome to the NCERT Solutions for Class 12 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 3: What is the shortest wavelength present in the Paschen series of spectral lines?....
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