Determine the series limit of Balmer, Paschen, and Pfund series, given the limit for Lyman series is 912 Å.

#### Solution

**Data: **`lambda_("L"∞)` = 912 Å

For hydrogen spectrum, `1/lambda = "R"_"H"(1/"n"^2 - 1/"m"^2)`

∴ `1/lambda_("L"∞) = "R"_"H"(1/1^2 - 1/∞) = "R"_"H"` ...(1)

as n = 1 and m = ∞

`1/lambda_("B"∞) = "R"_"H"(1/4 - 1/∞) = "R"_"H"/4` .....(2)

as n = 2 and m = ∞

`1/lambda_("Pa"∞) = "R"_"H"(1/9 - 1/∞) = "R"_"H"/9` .....(3)

as n = 3 and m = ∞

`1/lambda_("Pf"∞) = "R"_"H"(1/25 - 1/∞) = "R"_"H"/25` ......(4)

as n = S and m = ∞

From Eqs. (1) and (2), we get,

`lambda_("B"∞)/lambda_("L"∞) = "R"_"H"/("R"_"H"//4) = 4`

∴ `lambda_("B"∞) = 4 lambda_("L"∞) = (4)(912)` = 3648 Å

This is the series limit of the Balmer series.

rom Eqs. (1) and (3), we get,

`lambda_("Pa"∞)/lambda_("L"∞) = "R"_"H"/("R"_"H"//9)` = 9

∴ `lambda_("Pa"∞) = 9lambda_("L"∞) = (9)(912)` = 8208 Å

This is the series limit of the Paschen series.

From Eqs. (1) and (4), we get,

`lambda_("Pf"∞)/lambda_("L"∞) = "R"_"H"/("R"_"H"//25)` = 25

∴ `lambda_("Pf"∞) = 25 lambda_("L"∞) = (25)(912)` = 22800 Å