Wavelength of Photon Emitted During Electron Transition
Physics · Atoms · NEET
When an electron jumps from a higher orbit (n2) to a lower orbit (n1), it releases the extra energy as one photon. The wavelength of that photon comes from the Rydberg formula: 1/lambda = R (1/n1^2 - 1/n2^2), where R = 1.097 x 10^7 per metre. Memory hook: "energy out equals photon in" - bigger energy gap means shorter wavelength.
When the electron drops from a higher orbit n2 to a lower orbit n1, the energy difference leaves as one photon. Use the Rydberg formula with the lower orbit n1 first to find its wavelength.
Your doubts, answered
How do I find the wavelength of the photon when an electron jumps down?
First find the energy gap. The emitted photon energy equals E(n2) - E(n1) = 13.6 (1/n1^2 - 1/n2^2) eV, where n1 is the lower orbit. Then use the Rydberg form directly: 1/lambda = R (1/n1^2 - 1/n2^2) with R = 1.097 x 10^7 per metre. For example, for 3 to 2: 1/lambda = R (1/4 - 1/9) = R (5/36), which gives lambda about 656 nm (the red H-alpha line).
Which value goes first, n1 or n2, inside the Rydberg formula?
Always put the smaller orbit number (n1, the lower level) first so the bracket stays positive. So it is (1/n1^2 - 1/n2^2), not the other way. For emission the electron ends at n1 (the lower level) and starts at n2 (the higher level). If you flip them you get a negative number, which is your signal to swap them back.
Why is the wavelength shorter when the electron makes a bigger jump?
A bigger jump releases more energy. Photon energy E = hc/lambda, so energy and wavelength are inversely related. More energy means smaller lambda (shorter wavelength). That is why 6 to 2 gives a shorter wavelength than 3 to 2, even though both end at n1 = 2.
Is the energy gap the same as the wavelength?
No. The energy gap (in eV or joules) is what the atom loses. The wavelength is the property of the single photon that carries that energy away. They are linked by E = hc/lambda. A common quick conversion for NEET: lambda (in nm) = 1240 / E (in eV) when E is the photon energy in electron volts.
How do I compare wavelengths of two different transitions without a calculator?
Take the ratio. Since 1/lambda is proportional to (1/n1^2 - 1/n2^2), the ratio lambda_A / lambda_B equals the inverse ratio of these brackets. For 3 to 2 the bracket is 5/36; for 4 to 3 it is 7/144. So lambda(4 to 3) / lambda(3 to 2) = (5/36) / (7/144) = 20/7. This is the exact 2016 NEET trick.
⚠️ The NEET trap ✗ Students think a bigger orbit number means a longer wavelength, so they guess 4 to 3 gives a smaller wavelength than 3 to 2 and pick the wrong fraction like 9/16. ✓ Compute the brackets: 3 to 2 gives R(5/36) and 4 to 3 gives R(7/144). The 4 to 3 gap is smaller, so its wavelength is longer: lambda(4 to 3) = (20/7) lambda, which is bigger than lambda. Always take the ratio of (1/n1^2 - 1/n2^2), never guess from the orbit numbers alone. 🧠 The 3 to 2 vs 4 to 3 ratio trap
Real NEET questions
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: For 3 to 2: 1/lambda = R(1/2^2 - 1/3^2) = R(1/4 - 1/9) = R(5/36). For 4 to 3: 1/lambda' = R(1/3^2 - 1/4^2) = R(1/9 - 1/16) = R(7/144). Divide to get the ratio: lambda'/lambda = (5/36) / (7/144) = (5/36) x (144/7) = 720/252 = 20/7. So lambda' = (20/7) lambda.
2024
Match the hydrogen transitions (all ending at n1 = 2) with their wavelengths in nm. (A) 3 to 2 (B) 4 to 2 (C) 5 to 2 (D) 6 to 2; List II: (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: All transitions end at n1 = 2 (Balmer series). A larger jump gives a larger energy gap and a shorter wavelength. So order by jump size: (A) 3 to 2 is the smallest jump, longest wavelength 656.3 nm (III). (B) 4 to 2 = 486.1 nm (IV). (C) 5 to 2 = 434.1 nm (II). (D) 6 to 2 is the largest jump, shortest wavelength 410.2 nm (I). Answer: A-III, B-IV, C-II, D-I.
Solved Atoms NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the formula for the wavelength of a photon emitted in an electron transition?
1/lambda = R (1/n1^2 - 1/n2^2), where R = 1.097 x 10^7 per metre, n1 is the lower orbit and n2 is the higher orbit. This is the Rydberg formula for hydrogen.
What is the longest wavelength line in the Balmer series?
The 3 to 2 transition (H-alpha), about 656.3 nm, in the red part of visible light. It is the smallest jump in the Balmer series, so it has the smallest energy gap and the longest wavelength.
How is photon energy related to wavelength?
E = hc/lambda. In NEET-friendly units, lambda (nm) = 1240 / E (eV). So a 10.2 eV photon has a wavelength of about 122 nm, which is the Lyman-alpha line (2 to 1 transition).
Does the emitted photon wavelength depend on which series it belongs to?
Yes. The series is fixed by the final orbit n1: Lyman ends at n1 = 1 (ultraviolet), Balmer at n1 = 2 (visible), Paschen at n1 = 3 (infrared). Within a series, the wavelength shortens as n2 increases.
Can I use energy levels instead of the Rydberg formula?
Yes. Photon energy = E(n2) - E(n1) = 13.6 (1/n1^2 - 1/n2^2) eV. Then lambda (nm) = 1240 / E (eV). Both methods give the same answer; use whichever the question data suits.
