Physics, asked by avbsa21035shubhangi, 20 days ago

The wavelength of the first line of Lyman series of hydrogen atom is 1215 Å. Calculate the wavelength of the second line of the series, and the series limit​

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Answered by yashurajput881355
0

Explanation:

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Class 12

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>>If the first line of Lyman series has a

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If the first line of Lyman series has a wavelength 1215.4

A

˚

, the first line of Balmer series is approximately

Medium

Solution

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Correct option is C)

Wavelength of first line in Lyman series λ

L

=

4R

3

3R

4

=1215.4A

o

We get R=

3×1215.4

4

Wavelength of first line in Balmer series λ

B

=

5R

36

⟹ λ

B

=

5×4

36×3×1215.4

=6563A

o

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