Spectral Series of Hydrogen: Lyman, Balmer, Paschen, Brackett, Pfund
Physics · Atoms · NEET
The hydrogen spectrum has 5 named series based on the final orbit (n1) the electron falls to: Lyman (n1=1, ultraviolet), Balmer (n1=2, visible), Paschen (n1=3, infrared), Brackett (n1=4, infrared) and Pfund (n1=5, far infrared). Memory hook: "Little Boys Prefer Bright Paper" = Lyman, Balmer, Paschen, Brackett, Pfund for n1 = 1, 2, 3, 4, 5. Only the Balmer series lies in the visible region.
Energy level diagram of hydrogen: each series is named by the final orbit the electron falls to. Lyman (to n=1) is ultraviolet, Balmer (to n=2) is visible, Paschen and Brackett (to n=3, n=4) are infrared. Pfund ends at n=5 in far infrared.
Your doubts, answered
How do I know which series a spectral line belongs to?
Look only at the FINAL orbit n1 (the lower level the electron lands on). n1=1 is Lyman, n1=2 is Balmer, n1=3 is Paschen, n1=4 is Brackett, n1=5 is Pfund. The starting orbit n2 just tells you which line within that series (first line, second line, and so on). So a jump from n=5 to n=2 is a Balmer line because it ends at n1=2, not a Pfund line.
Which hydrogen series can we actually see with our eyes?
Only the Balmer series (n1=2) falls in the visible region (about 400 to 700 nm). Lyman (n1=1) is ultraviolet, so it is invisible. Paschen, Brackett and Pfund all lie in the infrared and far-infrared, also invisible. This is a very common NEET one-liner: 'visible series of hydrogen = Balmer'.
Why does the Lyman series have the shortest wavelengths of all series?
Lyman transitions end at n1=1, the lowest orbit, which is the largest energy gap. Wavelength is inversely related to energy (E = hc/lambda), so the biggest energy jumps give the smallest wavelengths. Because Lyman has the largest gaps, its lines have the shortest wavelengths and highest energy, placing them in the ultraviolet.
What is the difference between n1 and n2 in the Rydberg formula?
n1 is the lower (final) orbit and n2 is the higher (initial) orbit, with n2 greater than n1. In the formula 1/lambda = R(1/n1^2 - 1/n2^2), n1 fixes the SERIES and n2 fixes the LINE inside that series. For any series, the first line has n2 = n1+1 (longest wavelength) and the last line (series limit) has n2 = infinity (shortest wavelength).
Are Paschen, Brackett and Pfund really all infrared?
Yes. Paschen (n1=3) is near infrared, Brackett (n1=4) is infrared, and Pfund (n1=5) is far infrared. As n1 increases the energy gaps shrink, so wavelengths grow longer and move deeper into the infrared. A quick rule: n1=1 UV, n1=2 visible, n1>=3 infrared.
⚠️ The NEET trap ✗ A transition from n=6 to n=3 belongs to the Pfund series because the electron starts high up. ✓ It belongs to the Paschen series. The series is named by the FINAL orbit n1=3, not by the starting orbit. n1=3 is always Paschen no matter where the electron started. 🧠 Name the series by where the electron LANDS (n1), never by where it starts (n2).
Real NEET questions
NEET 2023 Phase 2
The wavelength of Lyman series of hydrogen atom appears in:
A · Ultraviolet region ✓
B · Infrared region
C · Visible region
D · Far infrared region
Solution: The Lyman series has n1=1, the lowest final orbit, so it involves the largest energy gaps and therefore the shortest wavelengths of all hydrogen series. Short wavelength plus high energy places these lines in the ultraviolet region. Correct option: A.
NEET 2016 Phase 1
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 m
B · 0.5 x 10^7 per m
C · 0.25 x 10^7 per m ✓
D · 2.5 x 10^7 per m
Solution: The 'last line' of a series is the series limit, the transition from n2 = infinity to the final orbit. For Balmer, n1 = 2. Wave number 1/lambda = R(1/n1^2 - 1/n2^2) = R(1/2^2 - 1/infinity) = R/4. With R = 10^7 per m, 1/lambda = 10^7 / 4 = 0.25 x 10^7 per m. Correct option: C.
NEET 2023 Phase 1
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 means the series limit (n2 = infinity). Balmer (n1=2): 1/lambda = R/2^2 = R/4. Brackett (n1=4): 1/lambda' = R/4^2 = R/16. Dividing, lambda'/lambda = (R/4)/(R/16) = 4, so lambda' = 4 lambda. Correct option: B.
Solved Atoms NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Five named series are studied at NEET level: Lyman, Balmer, Paschen, Brackett and Pfund, ending on orbits n1 = 1, 2, 3, 4 and 5 respectively. In principle more series exist (n1=6 and beyond) but they are not part of the syllabus.
What is the Rydberg formula for the hydrogen spectral series?
1/lambda = R(1/n1^2 - 1/n2^2), where R is the Rydberg constant (about 1.097 x 10^7 per metre), n1 is the final orbit and n2 is the initial orbit with n2 greater than n1. This one formula generates every line of every series.
Which series lies in the visible region and which in ultraviolet?
The Balmer series (n1=2) is the only one in the visible region. The Lyman series (n1=1) lies in the ultraviolet. Paschen, Brackett and Pfund all lie in the infrared and far infrared.
What is the difference between the first line and the last line of a series?
The first line uses the smallest jump, n2 = n1 + 1, and has the longest wavelength (smallest energy). The last line, called the series limit, uses n2 = infinity and has the shortest wavelength (largest energy) of that series.
Does the starting orbit decide the series name?
No. The series name depends only on the final orbit n1 where the electron lands. For example any transition ending at n1=2 is a Balmer line, regardless of whether it started from n=3, 5 or higher.
