# 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

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.

#### 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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#### APPEARS IN

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