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Answer in brief. Show that the frequency of the first line in the Lyman series is equal to the difference between the limiting frequencies of Lyman and the Balmer series. - Physics

Answer in Brief

Answer in brief.

Show that the frequency of the first line in the Lyman series is equal to the difference between the limiting frequencies of Lyman and the Balmer series.
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Solution

For the first line in the Lyman series,

`1/lambda_"L1" = "R"(1/1^2 - 1/2^2) = "R"(1 - 1/4) = "3R"/4`

∴ `"v"_"L1" = "c"/lambda_"L1" = "3Rc"/4`, where v denotes the frequency, c the speed of light in free space and R the Rydberg constant.

For the limit of the Lyman series,

`1/lambda_("L"∞) = "R"(1/1^2 - 1/∞)`= R

∴ `"v"_("L"∞) = "c"/(lambda_"L"∞) = "Rc"`

For the limit of the Balmer series,

`1/lambda_("B"∞) = "R"(1/2^2 - 1/∞) = "R"/4`

∴ `"v"_("B"∞) = "c"/(lambda_"B"∞) = "Rc"/4`

∴ `"v"_("L"∞) - "v"_("B"∞) = "Rc" - "Rc"/4 = "3Rc"/4 = "v"_"L1"`

Concept: Structure of Atoms and Nuclei
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Balbharati Physics 12th Standard HSC Maharashtra State Board
Chapter 15 Structure of Atoms and Nuclei
Exercises | Q 2.5 | Page 342
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