# The shortest and the longest wavelength in the Balmer series of the hydrogen spectrum are? Rydberg constant, RH = 109678 cm-1

By BYJU'S Exam Prep

Updated on: September 25th, 2023

(a) 911.7 Å and 1215.7 Å

(b) 3647 Å and 6565 Å

(c) 6565 Å and 3647 Å

(d) 911.7 Å and 6565 Å

The shortest and the longest wavelength in the Balmer series of the hydrogen spectrum are 3647 Å and 6565 Å. We know that for the Balmer series, n1 = 2

The energy difference between the two states exhibiting the transition should be at its maximum for the shortest wavelength in the Balmer series (also known as the series limit).

n2 = ∞

So 1/λ = RH [1/22 – 1/∞2] = RH/4

Substituting the values we get:

λ = 4/109678 = 3.647 x 10-5 cm

In simplification we get the:

= 3647 Å

The energy difference between the states showing the transition should be at its smallest for the longest wavelength in the Balmer series (i.e., the first line), i.e. n2 = 3

1/λ = RH [1/22 – 1/32] = 5/36RH

Substituting the values we get:

λ = 36/5 x 1/RH = 36/ (5 x 109678) = 6.565 x 10-5 cm

In simplification we get the:

= 6565 Å

### Hydrogen Spectrum

• A key piece of evidence for the quantized electronic structure of an atom is the hydrogen spectrum. As soon as an electric discharge passes through a gaseous hydrogen molecule, the molecule’s hydrogen atoms disintegrate.
• The energetically excited hydrogen atoms cause the emission of electromagnetic radiation, which is what happens.
• The hydrogen emission spectrum includes distinct frequency radiation. These radiation series bear the names of the researchers who first identified them.

Summary:

## The shortest and the longest wavelength in the Balmer series of the hydrogen spectrum are? Rydberg constant, RH = 109678 cm-1

The shortest and the longest wavelength in the Balmer series of the hydrogen spectrum are 3647 Å and 6565 Å. Balmer proposed the equation for relating the wave number of the emitted spectral lines and the involved energy shells in 1885, based on experimental observations.

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