de Broglie Wavelength and the Bohr Orbit Circumference (2πr = nλ)

Chemistry · Structure Of Atom · NEET

A Bohr orbit is stable only when a whole number of electron waves fit exactly around it. So the circumference of the orbit (2πr) equals n times the de Broglie wavelength: 2πr = nλ. This is why the electron does not lose energy and fall into the nucleus. Memory hook: "The wave must bite its own tail" — the wave must close on itself with no gap, like a ring made of n complete waves.
A Bohr orbit holds n whole de Broglie waves: 2πr = nλnucleusorbit (2πr)n whole waves (λ each), no gapmvr = nh/2π ⇔ λ = h/mv ⇔ 2πr = nλ
The electron behaves as a standing wave: the orbit circumference 2πr must equal n whole de Broglie wavelengths (nλ), which is the same as Bohr's rule mvr = nh/2π.

Your doubts, answered

Where does 2πr = nλ actually come from?

Start from Bohr's rule that angular momentum is quantised: mvr = nh/2π. de Broglie said every moving particle has a wavelength λ = h/mv, so mv = h/λ. Put mv = h/λ into Bohr's rule: (h/λ)·r = nh/2π. Cancel h from both sides: r/λ = n/2π, which rearranges to 2πr = nλ. So Bohr's angular-momentum rule and de Broglie's wave idea are the SAME statement.

What does 2πr = nλ physically mean?

2πr is the length once around the circular orbit (the circumference). nλ is n complete de Broglie waves laid end to end. The equation says the orbit length must hold a whole number of waves. If it held, say, 3.5 waves, the wave crest meeting a trough would cancel itself out — that orbit cannot exist. Only orbits that fit 1, 2, 3... whole waves are allowed.

How many de Broglie waves fit in the nth orbit?

Exactly n waves. In the 1st orbit (n=1) there is 1 whole wave, in the 2nd orbit (n=2) there are 2 whole waves, in the 3rd orbit 3 waves, and so on. The number of waves equals the orbit number n. This is a common one-line NEET fact.

How do I find the de Broglie wavelength of an electron in the nth Bohr orbit?

Use λ = 2πr/n. First find the orbit radius with rₙ = a₀·n²/Z (a₀ = 52.9 pm = 0.529 Å). Then divide the circumference 2πrₙ by n. Example for n=2 in hydrogen (Z=1): r₂ = 52.9 × 4 = 211.6 pm, so λ = 2π(211.6)/2 = 211.6π pm.

Does λ get bigger or smaller in higher orbits?

It gets BIGGER. For hydrogen, rₙ grows as n², so 2πrₙ grows as n². Dividing by n leaves λ ∝ n. So the wavelength grows in proportion to n: λ(n=2) is twice λ(n=1), λ(n=3) is three times, and so on. Radius grows faster (n²) but wavelength grows only as n.

Is 2πr = nλ the same as mvr = nh/2π?

Yes — they are two forms of one idea. mvr = nh/2π is Bohr's postulate (came first, was assumed). 2πr = nλ is de Broglie's explanation (came later, explains WHY the postulate is true). One turns into the other the moment you substitute λ = h/mv. NEET may ask either form.

⚠️ The NEET trap
For the 2nd Bohr orbit of hydrogen students compute λ = 2πr but forget to divide by n, getting 2π × 211.6, or they use r = 52.9 pm (the n=1 radius) instead of r = 211.6 pm.
λ = 2πrₙ / n. For n=2: rₙ = a₀n²/Z = 52.9 × 4 = 211.6 pm, so λ = 2π(211.6)/2 = 211.6π pm. The answer keeps the π — do NOT multiply π out unless a number is asked.
🧠 Two rules in one problem: (1) radius uses n² , (2) wavelength divides by n. Miss either and the answer is wrong.

Real NEET questions

NEET 2019 (Odisha)

In a hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is: [Bohr radius a₀ = 52.9 pm]

A · 211.6 pm
B · 211.6π pm
C · 52.9π pm
D · 105.8 pm
Solution: Bohr quantisation with de Broglie gives nλ = 2πrₙ. Radius: rₙ = a₀·n²/Z = 52.9 × 2²/1 = 211.6 pm. Wavelength: λ = 2πrₙ/n = 2π(211.6)/2 = 211.6π pm. Keep the π. Answer: (B).
NEET 2017 / 2018

Which one is the wrong statement?

A · de Broglie's wavelength is λ = h/mv, where m = mass and v = velocity of the particle
B · The uncertainty principle is ΔE·Δt ≥ h/4π
C · Half-filled and fully filled orbitals have greater stability
D · The energy of the 2s orbital is less than the energy of the 2p orbital in hydrogen-like atoms
Solution: In single-electron (hydrogen-like) atoms, energy depends only on n, so 2s and 2p are degenerate (equal energy) — statement (D) is wrong. The de Broglie relation λ = h/mv (option A) is correct and is the basis of 2πr = nλ. Answer: (D).

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Frequently asked

What is the formula linking de Broglie wavelength and Bohr orbit?

2πr = nλ, where r is the orbit radius, n is the orbit number (1, 2, 3...), and λ = h/mv is the de Broglie wavelength of the electron. It says the orbit circumference holds exactly n whole electron waves.

Why must the number of waves be a whole number?

A wave going once around must join back to itself smoothly. If a fraction of a wave were left over, the wave would meet itself out of step and cancel by destructive interference. Only whole numbers of waves survive, which is why n = 1, 2, 3... and orbits are quantised.

How does this remove a drawback of Bohr's model?

Bohr just assumed mvr = nh/2π without a reason. de Broglie's wave idea gives the reason: the electron is a standing wave around the orbit. So 2πr = nλ turns an assumption into a physical picture, which is a common NEET conceptual question.

Does 2πr = nλ work for any atom?

The relation itself holds for any single-electron (hydrogen-like) species. Only the radius changes: rₙ = 0.529 × n²/Z Å. Put that radius in and use λ = 2πrₙ/n.