The Rydberg formula gives the wave number (1/lambda) of any hydrogen spectral line: 1/lambda = R(1/n1^2 - 1/n2^2), where n1 is the lower orbit, n2 is the higher orbit, and R = 1.097 x 10^7 per metre. Wave number just means "how many waves fit in 1 metre," so its unit is per metre (m^-1), not metre. Memory hook: "small-square minus big-square" - always put the smaller orbit number (n1) first so the answer stays positive.
The Rydberg formula gives the wave number (1/lambda) of a hydrogen line; always put the smaller orbit as n1, and invert the result to get wavelength.
Your doubts, answered
Is wave number the same as wavelength?
No, they are opposites. Wave number is 1/lambda, meaning the number of full waves in one metre, with unit per metre (m^-1). Wavelength lambda is the length of one wave, with unit metre (m). The Rydberg formula directly gives wave number. To get wavelength, take the reciprocal: lambda = 1 / (wave number). Students lose marks by writing the wave number value as the wavelength.
What are n1 and n2 in the Rydberg formula?
n1 is the lower orbit (the final orbit the electron lands on) and n2 is the higher orbit (the initial orbit the electron jumps from). Always n1 is smaller than n2. For emission, the electron falls from n2 down to n1. Example: for the Balmer H-alpha line, the electron goes from n2 = 3 to n1 = 2, so you use 1/lambda = R(1/2^2 - 1/3^2).
What is the unit and value of the Rydberg constant R?
R = 1.097 x 10^7 per metre (m^-1). In many NEET numericals it is rounded to 10^7 m^-1 to make the arithmetic clean. Because R already has the unit m^-1, the whole right side of the formula has unit m^-1, which correctly matches wave number. If a question gives R in cm^-1 (about 1.097 x 10^5 cm^-1), your answer will come out in cm^-1 instead.
Why does the Rydberg formula always give a positive answer?
Because you write the smaller orbit as n1. Since 1/n1^2 is bigger than 1/n2^2 (smaller number squared gives a larger fraction), the bracket (1/n1^2 - 1/n2^2) is positive. Wave number and wavelength are physical lengths, so they must be positive. If you accidentally get a negative value, you swapped n1 and n2.
How do I get wavelength once I have the wave number?
Just flip it: lambda = 1 / (wave number). If 1/lambda = 0.25 x 10^7 m^-1, then lambda = 1 / (0.25 x 10^7) = 4 x 10^-7 m = 400 nm. Keep the powers of ten together to avoid slips. Do not forget to invert - this single step is the most common mistake in spectral-line problems.
Does the Rydberg formula work for He+ and Li2+?
Yes, with one change. For hydrogen-like ions you add Z^2: 1/lambda = R Z^2 (1/n1^2 - 1/n2^2), where Z is the nuclear charge (He+ has Z = 2, Li2+ has Z = 3). For plain hydrogen Z = 1, so it disappears. This Z^2 factor makes the same transition give a much shorter wavelength in heavier ions.
⚠️ The NEET trap ✗ For the last line of the Balmer series, students put n2 = 3 (the next orbit) and compute 1/lambda = R(1/4 - 1/9). ✓ The last line (series limit) is the transition from n2 = infinity to n1 = 2, so 1/lambda = R(1/4 - 0) = R/4 = 0.25 x 10^7 m^-1. 'Last line' means the shortest wavelength, which comes from the largest jump (n2 = infinity), not the first jump. 🧠 Last line = infinity jump. First line = nearest orbit. Never mix them up.
Real NEET questions
2016
Given the value of the Rydberg constant is 10^7 m^-1, the wave number of the last line of the Balmer series in the hydrogen spectrum will be:
A · 0.025 x 10^4 m^-1
B · 0.5 x 10^7 m^-1
C · 0.25 x 10^7 m^-1 ✓
D · 2.5 x 10^7 m^-1
Solution: Step 1: The 'last line' of a series is the series limit, the transition from n2 = infinity to the lower orbit. For the Balmer series the lower orbit is n1 = 2. Step 2: Apply the Rydberg formula: 1/lambda = R(1/n1^2 - 1/n2^2) = R(1/2^2 - 1/infinity^2) = R(1/4 - 0) = R/4. Step 3: Substitute R = 10^7 m^-1: 1/lambda = 10^7 / 4 = 0.25 x 10^7 m^-1. 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: Step 1: For 3 to 2: 1/lambda = R(1/2^2 - 1/3^2) = R(1/4 - 1/9) = R(9 - 4)/36 = R(5/36). Step 2: For 4 to 3: 1/lambda' = R(1/3^2 - 1/4^2) = R(1/9 - 1/16) = R(16 - 9)/144 = R(7/144). Step 3: Take the ratio lambda'/lambda = (1/lambda) / (1/lambda') = (5/36) / (7/144) = (5/36) x (144/7) = 720/252 = 20/7. Step 4: So lambda' = (20/7) lambda. Answer: C.
2024
Match List I (transitions in hydrogen) with List II (wavelengths in nm): (A) n2=3 to n1=2, (B) n2=4 to n1=2, (C) n2=5 to n1=2, (D) n2=6 to n1=2; (I) 410.2, (II) 434.1, (III) 656.3, (IV) 486.1.
A · A-III, B-IV, C-II, D-I ✓
B · A-IV, B-III, C-I, D-II
C · A-I, B-II, C-III, D-IV
D · A-II, B-I, C-IV, D-III
Solution: Step 1: All transitions end at n1 = 2, so this is the Balmer series. Step 2: A bigger jump (larger n2) means larger energy and therefore shorter wavelength. Step 3: Order the jumps: 3 to 2 is the smallest jump so it has the longest wavelength = 656.3 nm (III); 4 to 2 = 486.1 nm (IV); 5 to 2 = 434.1 nm (II); 6 to 2 is the largest jump so shortest wavelength = 410.2 nm (I). Step 4: Match: A-III, B-IV, C-II, D-I. Answer: A.
Solved Atoms NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the Rydberg formula for the hydrogen spectrum?
It is 1/lambda = R(1/n1^2 - 1/n2^2), where 1/lambda is the wave number of the emitted line, R = 1.097 x 10^7 m^-1 is the Rydberg constant, n1 is the lower orbit and n2 is the higher orbit.
What is wave number in physics?
Wave number is the number of complete waves that fit in one metre. It equals 1/lambda and has the unit per metre (m^-1). A larger wave number means a shorter wavelength.
What is the value of the Rydberg constant?
The Rydberg constant R = 1.097 x 10^7 per metre (m^-1). Many NEET problems round it to 10^7 m^-1 for simpler calculation.
How do you find the shortest wavelength of a series using the Rydberg formula?
Use n2 = infinity (the series limit), so 1/lambda = R/n1^2. For example, the shortest Balmer wavelength has n1 = 2, giving 1/lambda = R/4. The shortest wavelength always comes from the largest jump.
Why is the smaller orbit written as n1 in the Rydberg formula?
So that the bracket stays positive. Since 1/n1^2 is larger than 1/n2^2 when n1 is smaller, the difference is positive, and wave number must be positive because it is a physical quantity.
