Balmer Series of Hydrogen Spectrum Explained

Physics · Atoms · NEET

The Balmer series is the set of hydrogen spectral lines produced when an electron jumps DOWN to the second energy level (n = 2) from any higher level (n = 3, 4, 5...). These lines fall in the VISIBLE region, so Balmer is the only hydrogen series your eye can see. Memory hook: "B for Balmer, B for the eye-Ball you can see" - lands on n = 2, visible light.
Balmer Series: electrons falling to n = 2n=2n=3n=4n=5n=6H-alpha 656nmH-beta 486nmH-gamma 434nmVisiblelight
Balmer series: an excited electron drops from n = 3, 4, 5... down to n = 2. Each jump releases a visible photon. The smallest jump (3 to 2) gives red H-alpha at 656 nm (longest wavelength); larger jumps give shorter, bluer wavelengths.

Your doubts, answered

Is the Balmer series visible or in the ultraviolet region?

Balmer lines are in the VISIBLE region (about 400 to 700 nm). This is the one hydrogen series your eye can actually see. Only the far end of the series (the series limit near 365 nm) slips just into the near-ultraviolet. Do not confuse it with the Lyman series, which is fully ultraviolet.

Why does the Balmer series end at n = 2 and not n = 1?

Each series is named by the LOWER level the electron lands on. Balmer is defined as all jumps that finish at n = 2. Jumps that finish at n = 1 form a different series (Lyman). So n = 2 is the fixed floor for Balmer; the upper level n2 can be 3, 4, 5 and so on.

What is the longest wavelength line of the Balmer series?

The longest wavelength comes from the SMALLEST energy jump, which is n = 3 to n = 2. This is the H-alpha line at 656.3 nm (red light). Rule: smallest jump = smallest energy = longest wavelength = first line of the series.

What is the shortest wavelength of the Balmer series?

The shortest wavelength is the series limit, from n = infinity to n = 2. Using 1/lambda = R(1/4 - 0) = R/4, with R = 1.097 x 10^7 per metre, lambda = about 365 nm. This is the largest energy jump, so shortest wavelength and the last line of the series.

What is the difference between the Balmer and Lyman series?

Balmer lands on n = 2 and is visible; Lyman lands on n = 1 and is ultraviolet. Lyman jumps involve bigger energy gaps (electron falls all the way to the ground state), so Lyman photons carry more energy and have shorter wavelengths than Balmer photons.

Which Balmer line is the red H-alpha line?

The n = 3 to n = 2 transition gives H-alpha at 656.3 nm, which appears red. The next lines are H-beta (n=4 to 2, 486.1 nm, blue-green), H-gamma (n=5 to 2, 434.1 nm, violet) and H-delta (n=6 to 2, 410.2 nm, violet).

⚠️ The NEET trap
Students plug n1 = 1 (the ground state) into the Rydberg formula for the Balmer series, or they call Balmer an ultraviolet series like Lyman.
For Balmer ALWAYS use n1 = 2 as the lower level and n2 = 3, 4, 5... as the upper level. Balmer is VISIBLE. Only Lyman uses n1 = 1 and only Lyman is fully ultraviolet.
🧠 Balmer = 2 = eye-ball you can see. Lyman = 1 = you can't see (UV).

Real NEET questions

NEET 2016

Given the value of Rydberg constant is 10^7 per metre, the wave number of the last line of the Balmer series in hydrogen spectrum will be:

A · 0.025 x 10^4 per metre
B · 0.5 x 10^7 per metre
C · 0.25 x 10^7 per metre
D · 2.5 x 10^7 per metre
Solution: The 'last line' of a series is its series limit, the transition n = infinity to n = 2 for Balmer. Wave number = 1/lambda = R(1/n1^2 - 1/n2^2) = R(1/2^2 - 1/infinity^2) = R/4. With R = 10^7 per metre: wave number = 10^7 / 4 = 0.25 x 10^7 per metre. Correct option is C.
NEET 2023

In hydrogen spectrum, the shortest wavelength in the Balmer series is lambda. The shortest wavelength in the Brackett series is:

A · 2 lambda
B · 4 lambda
C · 9 lambda
D · 16 lambda
Solution: Shortest wavelength = series limit (upper level n = infinity). Balmer (infinity to 2): 1/lambda = R/2^2 = R/4. Brackett (infinity to 4): 1/lambda' = R/4^2 = R/16. Divide: (1/lambda)/(1/lambda') = (R/4)/(R/16) = 4, so lambda' = 4 lambda. Correct option is B.

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Frequently asked

What is the Rydberg formula for the Balmer series?

1/lambda = R(1/2^2 - 1/n2^2), where n2 = 3, 4, 5... and R = 1.097 x 10^7 per metre. The lower level is fixed at n1 = 2 for every Balmer line.

How many lines does the Balmer series have?

In theory an infinite number, because n2 can be 3, 4, 5 and so on up to infinity. The lines crowd closer together as they approach the series limit near 365 nm.

Why can we see the Balmer series but not the others?

Balmer jumps end at n = 2, giving medium-sized energy gaps whose photons have wavelengths of about 400 to 700 nm, which is exactly the visible range. Lyman (ending at n = 1) is higher energy and ultraviolet; Paschen, Brackett and Pfund end at n = 3 or higher and are infrared.

Which is the first line of the Balmer series?

The first line is H-alpha, the n = 3 to n = 2 transition at 656.3 nm (red). It is the smallest energy jump in the series, so it has the longest wavelength.

Does the Balmer series appear in emission or absorption?

It can appear in both. In emission the electron falls from a higher level to n = 2 and gives out a photon. In absorption an electron already in n = 2 absorbs a photon and jumps up, so the same wavelengths appear as dark lines.