Question

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.

Answer 1

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"`