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

Numerical

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

Concept: Atomic Spectra
Is there an error in this question or solution?

#### APPEARS IN

Balbharati Physics 12th Standard HSC Maharashtra State Board
Chapter 15 Structure of Atoms and Nuclei
Exercises | Q 6 | Page 342