Lyman Series and Why It Lies in the Ultraviolet Region
Physics · Atoms · NEET
The Lyman series is the set of spectral lines emitted when an excited electron in a hydrogen atom jumps down to the ground state, n = 1 (from n = 2, 3, 4, ... to n = 1). Because the electron falls into the lowest, most tightly bound level, the energy gap is the largest of any series, so the photon energy is highest and the wavelength is shortest. That is why every Lyman line lies in the ultraviolet region (about 91.2 nm to 121.6 nm). Memory hook: "Lyman Lands Low at n=1, so it is the Loftiest energy and Lands in UV."
Every Lyman line comes from an electron dropping to n = 1. Larger jumps (from higher n) release more energy, giving shorter ultraviolet wavelengths; the shortest is the 91.2 nm series limit and the longest is 121.6 nm.
Your doubts, answered
Why does the Lyman series lie in the ultraviolet region and not the visible region?
All Lyman lines end at n = 1, the ground state, which is the deepest energy level (-13.6 eV). Any electron falling to n = 1 crosses a very large energy gap, so it releases a high-energy photon. High energy means short wavelength (E = hc/lambda). The Lyman wavelengths (91.2 nm to 121.6 nm) are all shorter than 400 nm, which is the lower edge of visible light, so they fall in the ultraviolet region. In contrast, the Balmer series ends at n = 2 (smaller gap), giving visible light.
For the Lyman series, what is the value of the final orbit n1?
For the Lyman series the final (lower) orbit is always n1 = 1. The initial orbit n2 can be 2, 3, 4, 5 and so on up to infinity. So in the Rydberg formula 1/lambda = R(1/1^2 - 1/n2^2), you always put n1 = 1. If you accidentally use n1 = 2, you get the Balmer series instead.
What are the shortest and longest wavelengths of the Lyman series?
Longest wavelength (first line, smallest jump) comes from n2 = 2 to n1 = 1: 1/lambda = R(1 - 1/4) = 3R/4, giving lambda = 4/(3R) = about 121.6 nm. Shortest wavelength (series limit, largest jump) comes from n2 = infinity to n1 = 1: 1/lambda = R(1 - 0) = R, giving lambda = 1/R = about 91.2 nm. Both values are in the ultraviolet region.
Is the Lyman series visible to the human eye?
No. The human eye only sees roughly 400 nm to 700 nm. Every Lyman line is between 91.2 nm and 121.6 nm, which is below 400 nm, so it is ultraviolet and invisible. Only the Balmer series has lines in the visible region.
How is the Lyman series different from the Balmer series?
Lyman lines end at n1 = 1 and lie in the ultraviolet; Balmer lines end at n1 = 2 and lie in the visible region. Lyman has larger energy gaps, so shorter wavelengths. A common way to remember the order of increasing wavelength (and decreasing energy) is Lyman, Balmer, Paschen, Brackett, Pfund, matching final orbits n1 = 1, 2, 3, 4, 5.
⚠️ The NEET trap ✗ Using n1 = 2 for the Lyman series (confusing it with Balmer), or thinking Lyman is the visible series because it is the most famous. ✓ Lyman always ends at n1 = 1 and is entirely ultraviolet. The visible series is Balmer (n1 = 2). Lyman = UV, Balmer = visible, Paschen and beyond = infrared. 🧠 If they ask 'which series is ultraviolet' the answer is Lyman; if they ask 'which is visible' the answer is Balmer. Do not swap them.
Real NEET questions
2023
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 corresponds to all transitions ending at n1 = 1, the ground state. This is the deepest level, so the energy gaps are the largest of any hydrogen series and the emitted photons carry the highest energy. High energy means the shortest wavelengths (91.2 nm to 121.6 nm), which are all below 400 nm. Therefore every Lyman line lies in the ultraviolet region. Answer: A.
2017
The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is:
A · 2
B · 1
C · 4 ✓
D · 0.5
Solution: The 'last line' of a series is its series limit, the transition from n2 = infinity. Balmer last line (infinity to n1 = 2): 1/lambda_B = R(1/2^2 - 0) = R/4, so lambda_B = 4/R. Lyman last line (infinity to n1 = 1): 1/lambda_L = R(1/1^2 - 0) = R, so lambda_L = 1/R. Ratio lambda_B / lambda_L = (4/R)/(1/R) = 4. Answer: C.
2016
If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength lambda. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be:
A · 16/25 lambda
B · 9/16 lambda
C · 20/7 lambda ✓
D · 20/13 lambda
Solution: Use 1/lambda = R(1/n1^2 - 1/n2^2). For 3 to 2: 1/lambda = R(1/4 - 1/9) = R(5/36). For 4 to 3: 1/lambda' = R(1/9 - 1/16) = R(7/144). Divide: (1/lambda)/(1/lambda') = (5/36)/(7/144) = (5/36) x (144/7) = 20/7. So lambda'/lambda = 20/7, giving lambda' = (20/7) lambda. Answer: C. This uses the same Rydberg reasoning that governs the Lyman series.
Solved Atoms NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
1/lambda = R(1/1^2 - 1/n2^2), where R is the Rydberg constant (about 1.097 x 10^7 per metre) and n2 = 2, 3, 4, ... up to infinity. The final orbit is fixed at n1 = 1.
Who discovered the Lyman series?
It is named after the American physicist Theodore Lyman, who observed these ultraviolet hydrogen lines around 1906 to 1914.
Why is the Lyman series the highest-energy series in hydrogen?
Because it ends at n = 1, the most tightly bound level (-13.6 eV). Electrons falling here cross the biggest energy gaps, so they emit the most energetic (shortest wavelength) photons of all the hydrogen series.
What is the series limit of the Lyman series?
The series limit is the shortest wavelength, from n2 = infinity to n1 = 1: lambda = 1/R = about 91.2 nm. It corresponds to the ionisation energy of hydrogen from the ground state (13.6 eV).
Is the Lyman series an emission or absorption series?
It can be both. In emission, electrons fall to n = 1 and give out UV photons. In absorption, ground-state atoms (already at n = 1) absorb the same UV wavelengths to jump up, producing dark Lyman lines in an absorption spectrum.
