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Question : 10 of 51
Marks:
+1,
-0
Solution:
From formula,
‌=R[‌−‌] For Balmer series,
n1=2 ∴‌‌ For longest wavelength, transition occurs from
n=3 to
n=2.
∴‌=R[‌−‌] for shortest wavelength transition occurs from
n=∞ to
n=2 ∴‌‌‌=R[‌−‌] ∴‌‌‌| λ‌longest ‌ |
| λ‌shorts ‌ |
=‌ For Paschen series,
n1=3 λ‌longest ‌ of Balmer
=‌ and
λ‌shortest ‌ of Paschen
=‌ Hence there is no overlap between the wavelength ranges of Balamer and Paschen series.
For Lyman series,
n1=1 ‌=R[‌−‌] Also
‌=R ∴‌‌‌=‌[1−‌]⇒λ=‌ λ‌longest ‌ of Lyman
=‌ and
λ‌shortest ‌ of Balmer
=‌ Hence that wavelenght ranges of Lyman and Balmer series do not overlap.
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