SELECT * FROM question_mgmt as q WHERE id=2764 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=8 AND chapterId=126 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 12th physics Atoms what is the shortest wavelength present in the pas

Question:

What is the shortest wavelength present in the Paschen series of spectral lines?

Answer:

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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