Physics · Atoms

de Broglie's Explanation of Bohr's Second Postulate

Bohr asserted that the angular momentum of the orbiting electron is an integral multiple of $h/2\pi$, but he offered no reason for the rule. In 1923 Louis de Broglie supplied one: the electron is a matter wave, and only orbits that hold a whole number of wavelengths form a stable standing wave. This page follows NCERT Class 12 §12.6 to show how the standing-wave condition $2\pi r_n = n\lambda$, combined with $\lambda = h/mv_n$, reproduces Bohr's quantisation $mv_nr_n = nh/2\pi$ — a recurring NEET theme.

The Unexplained Postulate

In Bohr's model of the hydrogen atom, three postulates carry the entire structure. The first fixes the existence of stable, non-radiating orbits. The third governs the emission and absorption of photons during transitions. Between them sits the second postulate, which NCERT calls "perhaps the most puzzling" of the three. It states that the electron revolves only in those orbits for which the angular momentum is an integral multiple of $h/2\pi$:

$$L_n = \frac{nh}{2\pi}, \qquad n = 1, 2, 3, \dots$$

This single condition is what converts a continuous family of classical orbits into a discrete ladder of allowed radii and energies. The radius of the $n$th orbit and its energy both follow from it. Yet Bohr provided no physical mechanism for the rule. The natural question — why should angular momentum take only these particular values, and no values in between — remained open for a decade.

The dissatisfaction was not merely aesthetic. Bohr fixed the integer $h/2\pi$ by demanding that his model reproduce the already-known wavelengths of the hydrogen spectrum; the quantisation rule was reverse-engineered to fit experiment rather than deduced from any deeper principle. A postulate chosen this way works wherever it was fitted but carries no guarantee of being correct, and it gives no insight into why nature should single out integer multiples of $h/2\pi$ and forbid everything in between. This is exactly the sort of unexplained quantisation that an honest physical theory should account for, and it is the gap de Broglie set out to close.

NEET Trap

Bohr stated the rule; de Broglie explained it

A common confusion is treating the quantisation of angular momentum as something de Broglie discovered. Bohr postulated $L_n = nh/2\pi$ in 1913 as an unproven assumption. de Broglie's 1923 contribution was to justify that postulate using the wave nature of the electron. The equation is Bohr's; the reasoning behind it is de Broglie's.

If a question asks "who proposed $L = nh/2\pi$" the answer is Bohr; if it asks "who explained why it holds," the answer is de Broglie.

Electron as a Matter Wave

The resolution rests on the de Broglie hypothesis introduced in the dual-nature chapter: material particles, including electrons, possess a wave nature. A particle of momentum $p$ has an associated wavelength

$$\lambda = \frac{h}{p}.$$

For an electron moving much slower than light, the momentum is simply $p = mv_n$, so the de Broglie wavelength of an electron in the $n$th orbit is $\lambda = h/mv_n$. The reality of this wave was not a theoretical convenience: C. J. Davisson and L. H. Germer experimentally verified the wave nature of electrons in 1927 through electron diffraction. de Broglie's insight was to take this established wave nature seriously inside the atom itself.

Louis de Broglie argued that the electron in its circular Bohr orbit must be treated not as a tiny planet but as a particle wave wrapped around the nucleus. The behaviour of such a wave on a closed loop is governed by the same physics that decides which notes a plucked string can sustain.

A matter wave wrapped around a circular orbit An electron treated as a wave forms a closed loop around the nucleus; the orbit is the dashed circle and the wavy line is the electron wave riding along it. + orbit r electron wave
The orbiting electron is viewed as a matter wave of wavelength $\lambda = h/mv_n$ riding on the circular path of radius $r_n$. Only specific wavelengths let the wave join smoothly onto itself.

Standing Waves on a Circular Orbit

The string analogy is exact and is the heart of the argument. When a string is plucked, a vast number of wavelengths are excited, but only certain ones survive. The survivors are those that have nodes at the ends and form a standing wave; the rest interfere with themselves on reflection and their amplitudes quickly drop to zero. In NCERT's words, standing waves on a string form when the total distance travelled by the wave down the string and back is one wavelength, two wavelengths, or any integral number of wavelengths.

Translate this to a circular orbit. The electron wave does not reflect off ends; instead it travels around the loop and must meet itself. The total distance the wave covers in one trip around the $n$th orbit is the circumference, $2\pi r_n$. For the wave to be a self-consistent standing wave — for crest to land on crest after a full circuit — that circumference must contain a whole number of de Broglie wavelengths:

$$2\pi r_n = n\lambda, \qquad n = 1, 2, 3, \dots \tag{12.12}$$

If the circumference were instead, say, $3.5\lambda$, the wave returning to its starting point would arrive out of phase, crest meeting trough, and the loop would cancel itself within a few circuits. Such an orbit cannot persist. Only resonant orbits, holding an integral number of wavelengths, support a stable standing wave.

A standing wave with four de Broglie wavelengths on a circular orbit A closed wavy loop showing four full wavelengths fitting exactly around the circular orbit, the n equals four case from NCERT Figure 12.8. n = 4 2πr₄ = 4λ
NCERT Figure 12.8: a standing particle wave on a circular orbit for $n = 4$, where four de Broglie wavelengths fit into the circumference. The dark dots mark nodes where successive loops join. An orbit holding a non-integer number of wavelengths interferes destructively and is forbidden.
Build the foundation

The wavelength $\lambda = h/mv$ used here is the same matter wave that produces the discrete hydrogen line spectra through quantised energy levels.

Deriving the Quantisation Condition

The standing-wave condition contains Bohr's postulate inside it. Begin with the resonance requirement and substitute the de Broglie wavelength. From Eq. (12.12),

$$2\pi r_n = n\lambda.$$

With $\lambda = h/mv_n$, this becomes

$$2\pi r_n = \frac{nh}{mv_n} \qquad \Longrightarrow \qquad m v_n r_n = \frac{nh}{2\pi}.$$

The left side $m v_n r_n$ is precisely the angular momentum $L_n$ of the electron in a circular orbit. The derivation therefore yields

$$L_n = m v_n r_n = \frac{nh}{2\pi},$$

which is exactly the quantum condition Bohr proposed for the angular momentum. What Bohr had to assume, de Broglie produced as a consequence of the electron behaving as a wave. The discrete orbits and energy levels of hydrogen are, in this view, the resonant modes of a standing electron wave; only those modes can persist.

It is worth pausing on the direction of logic, because the algebra can hide it. Bohr started from $L = nh/2\pi$ and treated the integer $n$ as an unexplained quantum number. de Broglie starts one step earlier, from the physical demand that the electron wave be self-consistent around the loop, and the integer $n$ then emerges automatically as the count of wavelengths around the orbit — it is not inserted by hand. The factor $2\pi$ in $nh/2\pi$ likewise has a transparent origin here: it is simply the geometric factor that converts the orbit circumference $2\pi r_n$ into the radius $r_n$ when the de Broglie relation is substituted. A rule that looked arbitrary in 1913 becomes, in de Broglie's reading, the most natural condition imaginable for a confined wave.

StepRelationSource / meaning
Standing-wave condition2πrₙ = nλWhole number of wavelengths around circumference (Eq. 12.12)
de Broglie wavelengthλ = h/mvₙWave nature of the electron, $p = mv_n$
Substitute2πrₙ = nh/mvₙEliminate $\lambda$
Rearrangemvₙrₙ = nh/2πBohr's quantisation of angular momentum
NEET Trap

2πrₙ = nλ is the cause; mvr = nh/2π is the consequence

Students often memorise $mvr = nh/2\pi$ without seeing where it comes from. The logical chain is one-directional: the standing-wave requirement $2\pi r_n = n\lambda$ is the physical input, and only standing waves with a whole number of wavelengths survive. Plugging in $\lambda = h/mv_n$ then produces $mv_nr_n = nh/2\pi$. This is the reason angular momentum is quantised — not an independent fact to be memorised alongside it.

If asked "why is angular momentum quantised in the Bohr atom?", the correct answer is: only standing electron waves ($2\pi r_n = n\lambda$) can persist, and these force $L = nh/2\pi$.

Allowed Versus Forbidden Orbits

The standing-wave picture draws a sharp line between permitted and impossible orbits. An orbit is allowed only when its circumference equals an exact integer number of wavelengths. Any orbit whose circumference works out to a fractional number of wavelengths — $1.3\lambda$, $2.7\lambda$, $3.5\lambda$ — fails to close on itself; the wave interferes destructively with its own tail and dies out. There are no orbits "between" the allowed ones for the same reason there are no notes between the harmonics of a fixed string.

Allowed integer-wavelength orbit versus forbidden fractional-wavelength orbit Left circle shows a wave closing smoothly with a whole number of wavelengths and is labelled allowed; right circle shows a wave that fails to meet itself with a gap and is labelled forbidden. ALLOWED 2πr = nλ (integer) mismatch FORBIDDEN 2πr ≠ nλ (fractional)
An allowed orbit (left) holds a whole number of wavelengths, so the wave joins itself and reinforces. A forbidden orbit (right) holds a non-integer number; the returning wave is out of phase with its start, the loop interferes destructively, and the orbit cannot exist.

This is the deep content of the second postulate as de Broglie reframed it. The quantised electron orbits and energy states of hydrogen arise from the wave nature of the electron, and only resonant standing waves can persist. The integer $n$ that appears in $L_n = nh/2\pi$ is, physically, the number of de Broglie wavelengths fitting around the orbit.

Worked Numbers and Limits

Two relations from the Bohr model make the standing-wave picture quantitative. The allowed radius grows as $r_n = 0.529\,n^2$ Å, and combining the resonance condition with the orbit radius lets the de Broglie wavelength be written directly in terms of the orbit, since $2\pi r_n = n\lambda$ rearranges to $\lambda = 2\pi r_n / n$.

Worked Example

For the ground state of hydrogen ($n = 1$, $r_1 = 0.529$ Å), how many de Broglie wavelengths fit around the orbit, and what is the wavelength $\lambda$?

The number of wavelengths around the orbit equals $n$ by the standing-wave condition, so for $n = 1$ exactly one de Broglie wavelength fits around the first orbit. Its length follows from $2\pi r_1 = 1\cdot\lambda$:

$$\lambda = 2\pi r_1 = 2\pi (0.529 \times 10^{-10}\,\text{m}) \approx 3.32 \times 10^{-10}\,\text{m}.$$

So the matter wave of the ground-state electron has a wavelength of about 3.3 Å, exactly equal to the circumference of the first orbit — a single closed loop.

Worked Example

How does the de Broglie wavelength of the electron change as it moves to higher orbits, and how many wavelengths span the $n = 3$ orbit?

By the resonance condition the number of wavelengths around any orbit equals $n$, so the $n = 3$ orbit holds exactly three de Broglie wavelengths. The length of each follows from $\lambda = 2\pi r_n / n$ with $r_n = 0.529\,n^2$ Å:

$$\lambda = \frac{2\pi (0.529\,n^2\,\text{Å})}{n} = 2\pi (0.529)\,n\,\text{Å} \approx 3.32\,n\ \text{Å}.$$

For $n = 3$ this gives $\lambda \approx 9.97$ Å. The wavelength therefore grows in direct proportion to $n$: the electron in a higher orbit is a "longer", slower wave, consistent with $v_n \propto 1/n$ in the Bohr model. Three of these longer waves close the third orbit just as one short wave closes the first.

Significance and Quantum Mechanics

What de Broglie achieved is best stated precisely. He did not derive the Bohr model from scratch, and he did not remove its failures. He supplied the missing reason for one postulate: the quantisation of angular momentum is the resonance condition for a confined electron wave. That single reinterpretation transformed the second postulate from an arbitrary rule into a physically motivated requirement, and it pointed the way to a far deeper theory.

The standing-wave picture is nonetheless only a heuristic. It treats the electron wave as a one-dimensional loop riding on a sharply defined circular path of radius $r_n$ — but the wave nature of the electron is precisely what makes such a definite path untenable, since a wave spread around a loop cannot also be a point with a fixed position and a fixed momentum. The argument is suggestive and gives the right quantisation condition, yet it stops short of a complete description of the atom.

NCERT is explicit on this point: a thorough quantum-mechanical treatment of the hydrogen atom — replacing the orbiting-particle picture with a three-dimensional probability wave governed by the Schrödinger equation — was developed shortly afterward by Schrödinger and others, and it yields the energy levels of hydrogen without invoking Bohr's postulates at all. de Broglie's standing wave is best seen as the conceptual bridge between Bohr's quantised orbits and that full wave mechanics: enough to explain why $L = nh/2\pi$, but not the final word.

The argument also has firm limits, and NEET tests these boundaries. de Broglie's reasoning explains why angular momentum is quantised, but it inherits every restriction of the Bohr model. The model is valid only for hydrogenic, single-electron systems — hydrogen, He⁺, Li²⁺ — because it ignores electron–electron forces. It cannot be extended even to neutral helium, and it cannot explain the relative intensities of spectral lines. A complete account requires full quantum mechanics.

Quantity in nth orbitExpressionStanding-wave reading
Circumference2πrₙTotal path of the wave per circuit
Resonance condition2πrₙ = nλHolds exactly $n$ wavelengths
Angular momentumLₙ = nh/2πForced by the resonance
Radius (hydrogen)rₙ = 0.529 n² ÅAllowed standing-wave radii only
Wavelength in orbitλ = 2πrₙ / nRearranged resonance condition
Quick Recap

de Broglie and the second postulate in one screen

  • Bohr postulated $L_n = nh/2\pi$ (1913); de Broglie explained it (1923).
  • The electron is a matter wave with $\lambda = h/mv_n$; Davisson–Germer confirmed the wave nature in 1927.
  • A stable orbit must support a standing wave: $2\pi r_n = n\lambda$, an integer number of wavelengths.
  • Substituting $\lambda = h/mv_n$ gives $m v_n r_n = nh/2\pi$ — Bohr's quantisation condition.
  • Orbits with a fractional number of wavelengths interfere destructively and are forbidden.
  • The integer $n$ is the number of de Broglie wavelengths around the orbit.
  • The argument explains quantisation but keeps the Bohr model's limits: single-electron atoms only.

NEET PYQ Snapshot — de Broglie's Explanation of Bohr's Second Postulate

Real NEET questions tied to the matter wave, the standing-wave condition, and quantisation of angular momentum.

NEET 2025 · Q.27

De-Broglie wavelength of an electron orbiting in the $n = 2$ state of hydrogen atom is close to (Given Bohr radius = 0.052 nm).

  • (1) 2.67 nm
  • (2) 0.067 nm
  • (3) 0.67 nm
  • (4) 1.67 nm
Answer: (3) 0.67 nm

Orbit radius $r = 0.052\,n^2 = 0.052 \times 4 = 0.208$ nm. The standing-wave condition gives $2\pi r = n\lambda$, so $\lambda = 2\pi r / n = \pi r$ for $n = 2$. Thus $\lambda = 3.14 \times 0.208 \approx 0.65$ nm $\approx 0.67$ nm. This is a direct test of $2\pi r_n = n\lambda$.

NEET 2025 · Q.18

A particle of mass $m$ moves around the origin under a constant force $F$ pulling it towards the origin. Using the Bohr model, the radius of the $n$th orbit and the speed $v$ depend on $n$ as:

  • (1) $r \propto n^{4/3}$; $v \propto n^{-1/3}$
  • (2) $r \propto n^{1/3}$; $v \propto n^{1/3}$
  • (3) $r \propto n^{1/3}$; $v \propto n^{2/3}$
  • (4) $r \propto n^{2/3}$; $v \propto n^{1/3}$
Answer: (4)

Constant force gives $F = mv^2/r$, so $v^2/r$ is constant and $r \propto v^2$. The quantisation $L = mvr = nh/2\pi$ — the very condition de Broglie's standing wave produces — closes the system. Solving the two relations gives $v \propto n^{1/3}$ and $r \propto n^{2/3}$.

FAQs — de Broglie's Explanation of Bohr's Second Postulate

Quick answers to the questions NEET aspirants ask most about the standing electron wave.

What is de Broglie's explanation of Bohr's second postulate?
de Broglie argued that the orbiting electron is a matter wave. A wave can persist on a circular orbit only if it forms a standing wave, which requires the circumference to hold a whole number of wavelengths: 2πrn = nλ. Substituting the de Broglie wavelength λ = h/mvn gives mvnrn = nh/2π, exactly Bohr's quantisation condition for angular momentum. The orbit quantisation is therefore a resonance condition for the electron wave.
Why is the angular momentum of the electron quantised?
Because only those orbits survive in which the electron wave closes on itself as a standing wave. The standing-wave condition 2πrn = nλ allows only an integral number of wavelengths around the circumference. Using λ = h/mvn this becomes mvnrn = nh/2π, so the angular momentum is forced to be an integer multiple of h/2π. Orbits that do not fit a whole number of wavelengths produce waves that interfere destructively and cannot persist.
What does the equation 2πrn = nλ represent?
It is the standing-wave condition for an electron in the nth circular orbit. The total distance travelled by the wave around the orbit is its circumference, 2πrn. For a standing wave this distance must equal a whole number of de Broglie wavelengths, so 2πrn = nλ with n = 1, 2, 3, …. Figure 12.8 of NCERT shows the n = 4 case where four wavelengths fit the circumference.
Who proposed this explanation and in which year?
The French physicist Louis de Broglie proposed it in 1923, ten years after Bohr published his model in 1913. The underlying wave nature of the electron, λ = h/mv, was experimentally confirmed by C. J. Davisson and L. H. Germer in 1927.
Does de Broglie's argument remove all limitations of the Bohr model?
No. It explains why angular momentum is quantised, giving a physical basis for the second postulate, but the Bohr model still applies only to hydrogenic single-electron atoms, cannot handle helium or multi-electron atoms, and cannot explain the relative intensities of spectral lines. A complete description needs full quantum mechanics.