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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

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प्रश्न

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.
संक्षेप में उत्तर

उत्तर

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"`

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Structure of Atoms and Nuclei
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अध्याय 15: Structure of Atoms and Nuclei - Exercises [पृष्ठ ३४२]

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बालभारती Physics [English] 12 Standard HSC Maharashtra State Board
अध्याय 15 Structure of Atoms and Nuclei
Exercises | Q 2.5 | पृष्ठ ३४२

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