de Broglie's Explanation & Limitations of Bohr's Model

Why nh/2π? De Broglie's Standing-Wave Answer

Of Bohr's three postulates, the second is the most puzzling. Why should angular momentum come only in integral multiples of $h/2\pi$? Bohr offered no reason — it simply worked. The French physicist Louis de Broglie explained the puzzle in 1923, ten years after the model appeared.

His idea: the orbiting electron (whose wave nature Davisson and Germer would confirm in 1927) must be seen as a particle wave — and particle waves, like waves on a plucked string, form standing waves only under resonant conditions.

The string analogy

When a string is plucked, countless wavelengths are excited — but only those that fit (nodes at the ends, whole numbers of half-wavelengths) survive as standing waves; all others interfere destructively with themselves and die out.

For an electron in a circular orbit of radius $r_n$, the 'string' closes on itself: the wave must fit the circumference exactly:

$$\boxed{2\pi r_n = n\lambda}, \qquad n = 1, 2, 3, \ldots$$

(NCERT Fig. 12.8 shows n = 4: four de Broglie wavelengths wrapped around the orbit.)

Two lines of algebra

Substituting $\lambda = \dfrac{h}{p} = \dfrac{h}{mv_n}$:

$$2\pi r_n = \frac{nh}{mv_n} \quad\Rightarrow\quad mv_nr_n = \frac{nh}{2\pi}$$

This is exactly Bohr's second postulate. The quantised orbits exist because only resonant standing electron waves persist — quantisation is the wave nature of the electron in disguise.

The Honest Audit: Limitations of Bohr's Model

Bohr's model correctly predicts the gross features of hydrogenic atoms — especially the frequencies of emitted and absorbed radiation. But NCERT lists its limits plainly:

Limitation 1: Hydrogenic atoms only

The model works for hydrogenic (one-electron) atoms: H, He$^+$, Li$^{2+}$, … It cannot be extended even to helium (two electrons). The reason: Bohr's formulation includes only the nucleus-electron electrical force; in multi-electron atoms, each electron also interacts with all other electrons, and these electron-electron forces are comparable in magnitude to the nuclear pull (same order of charges and distances). Unlike the solar system — where planet-planet gravity is negligible next to the Sun's — the atom offers no such small parameter.

Limitation 2: No intensities

The model predicts the frequencies of hydrogen's lines but is silent on their relative intensities — why some transitions are strong and others weak. Experiment shows some transitions are 'more favoured' than others; Bohr's model cannot say why.

For complex atoms — and even for a full account of hydrogen — physics needed the radical new framework of quantum mechanics.

[NEET Important] The two limitations, verbatim, are a standard 2-marker: (i) fails for multi-electron atoms (electron-electron interactions omitted), (ii) cannot explain relative intensities of spectral lines.

Points to Ponder: The Deep Cuts — [JEE/NEET Extra Insight]

NCERT's closing reflections furnish the assertion-reason questions of tomorrow:

• Why angular momentum? Planck's constant h has the dimensions of angular momentum ([M L$^2$ T$^{-1}$]), and for circular orbits angular momentum is the natural variable — 'the second postulate is then so natural!'
• Orbits are not literal. Bohr's trajectory picture is inconsistent with the uncertainty principle; modern quantum mechanics replaces orbits with regions of high probability of finding the electron.
• Revolution frequency ≠ spectral frequency. Contrary to classical expectation, the frequency of the emitted line is $(E_i - E_f)/h$, not the orbital frequency — except for transitions between very large adjacent n, where the two coincide. (This is the correspondence principle: quantum results merge into classical ones at large quantum numbers.)
• One number isn't enough. Bohr's model carries only n; full quantum mechanics characterises states by four quantum numbers (n, l, m, s) — though for the pure Coulomb potential the energy still depends only on n.
• Why keep a 'wrong' model? Three postulates account for almost all gross features of hydrogen's spectrum; the model leans on familiar classical concepts; and it shows how a physicist may deliberately set aside difficulties to extract predictions — rationalising them later if experiment agrees.

[JEE Tip] The standing-wave condition doubles as a formula: the de Broglie wavelength in the nth orbit is $\lambda_n = \dfrac{2\pi r_n}{n} = \dfrac{2\pi a_0 n^2}{n} = 2\pi a_0 n$ — proportional to n. For n = 1: $\lambda_1 = 2\pi a_0 = 3.3 \times 10^{-10}$ m, exactly one circumference.

Solved Examples

Example 1: Deriving Bohr from de Broglie [Board Derivation]

Show that the standing-wave condition on a circular orbit yields Bohr's quantisation of angular momentum.

Solution:

1. Resonance condition: the electron wave must close smoothly on itself: $2\pi r_n = n\lambda$.
2. de Broglie: $\lambda = \dfrac{h}{mv_n}$ (speed well below c, so p = mv).
3. Substitute: $2\pi r_n = \dfrac{nh}{mv_n}$.
4. Rearrange: $mv_nr_n = \dfrac{nh}{2\pi}$ — Bohr's second postulate, derived rather than decreed. ∎

Example 2: Wavelength in the nth orbit [JEE Numerical]

Find the de Broglie wavelength of hydrogen's electron in the n = 1 and n = 4 orbits.

Solution:

1. From the fit condition: $\lambda_n = \dfrac{2\pi r_n}{n} = \dfrac{2\pi n^2 a_0}{n} = 2\pi n a_0$.
2. n = 1: $\lambda_1 = 2\pi \times 5.3 \times 10^{-11} = 3.3 \times 10^{-10}$ m.
3. n = 4: $\lambda_4 = 4\lambda_1 = 1.33 \times 10^{-9}$ m.
4. Takeaway: $\lambda_n \propto n$ — one wavelength fills the first orbit, four fill the fourth (NCERT Fig. 12.8's picture).

Example 3: Why helium defeats Bohr [Board Conceptual]

Explain precisely why Bohr's model cannot handle the helium atom.

Solution:

1. Helium has two electrons; each interacts with the nucleus AND with the other electron.
2. Bohr's formulation includes only the nucleus-electron force; it has no machinery for electron-electron forces.
3. These cannot be neglected: the charges and distances involved are of the same order, so the e-e force is comparable to the nuclear pull (NCERT Points to Ponder 4).
4. Contrast: the solar system survives a similar 'many-body' worry because planet-planet gravity is genuinely tiny next to the Sun's — no analogous smallness exists in the atom.

Example 4: Standing vs travelling waves [Conceptual]

Why do non-resonant electron wavelengths not correspond to allowed orbits?

Solution:

1. A wave that fails the fit condition returns to each point out of phase with itself after each circuit.
2. Successive circuits interfere destructively; the amplitude 'quickly drops to zero' (NCERT's string language).
3. Only wavelengths satisfying $2\pi r = n\lambda$ reinforce themselves — resonant standing waves persist, and these are precisely Bohr's orbits.

Example 5: Counting nodes on the n = 4 orbit [NEET Pattern]

How many de Broglie wavelengths fit around hydrogen's n = 4 orbit, and how many nodes does the standing wave have?

Solution:

1. Fit condition: $2\pi r_4 = 4\lambda$ — four wavelengths wrap the orbit (NCERT Fig. 12.8 exactly).
2. Nodes: a circular standing wave with n wavelengths has 2n nodes (each wavelength contributes two) — here 8 nodes.
3. Takeaway: 'number of wavelengths = n' is the picture exams draw; count nodes as 2n if asked.

Example 6: Ratio of de Broglie wavelengths [JEE Ratio]

Find the ratio of the electron's de Broglie wavelengths in the n = 2 and n = 3 orbits of hydrogen.

Solution:

1. Formula: $\lambda_n = 2\pi n a_0 \propto n$.
2. Ratio: $\lambda_2 : \lambda_3 = 2 : 3$.
3. Cross-check via speeds: $\lambda = h/mv_n$ and $v_n \propto 1/n$ → $\lambda \propto n$. ✔
4. Takeaway: two independent routes, one answer — the mark of a consistent model.

Example 7: The correspondence check [JEE Advanced Flavour]

NCERT states that for transitions n → n-1 with very large n, the emitted frequency approaches the orbital revolution frequency. Verify the scaling.

Solution:

1. Emitted frequency: $\nu = \dfrac{E_n - E_{n-1}}{h} \propto \left(\dfrac{1}{(n-1)^2} - \dfrac{1}{n^2}\right) = \dfrac{2n - 1}{n^2(n-1)^2} \approx \dfrac{2}{n^3}$ for large n.
2. Orbital frequency: $\nu_{orb} = \dfrac{v_n}{2\pi r_n} \propto \dfrac{1/n}{n^2} = \dfrac{1}{n^3}$ — with matching constants, the two coincide as n → ∞.
3. Takeaway: quantum → classical at large quantum numbers: Bohr's correspondence principle, NCERT Points to Ponder 6.

Example 8: What the modern picture keeps and discards [Conceptual]

According to NCERT's Points to Ponder, what happens to Bohr's orbits in modern quantum mechanics?

Solution:

1. The orbital picture is inconsistent with the uncertainty principle (exact r and v simultaneously is forbidden).
2. Orbits are replaced by regions where the electron may be found with large probability — probability clouds, not tracks.
3. What survives: the energy levels $E_n$ and the transition rule $h\nu = E_i - E_f$ — the experimentally tested content.
4. Takeaway: Bohr's numbers were right; his trajectories were scaffolding.

Example 9: Momentum from the fit condition [JEE Numerical]

Using only $2\pi r_n = n\lambda$ and $r_1 = a_0 = 5.3 \times 10^{-11}$ m, find the electron's momentum in hydrogen's ground state.

Solution:

1. Wavelength: $\lambda_1 = 2\pi a_0 = 3.33 \times 10^{-10}$ m.
2. de Broglie: $p = \dfrac{h}{\lambda} = \dfrac{6.63 \times 10^{-34}}{3.33 \times 10^{-10}}$.
3. Answer: $p \approx 2.0 \times 10^{-24}$ kg m/s.
4. Check: $p = mv = 9.1 \times 10^{-31} \times 2.18 \times 10^6 = 1.98 \times 10^{-24}$ kg m/s. ✔ Geometry and dynamics agree.

Example 10: Why gravity can't rescue Bohr for helium [Conceptual]

The solar system is a many-body system, yet Newtonian 'Bohr-style' analysis of each planet works well. Why does the analogous trick fail for helium's two electrons?

Solution:

1. Solar system: planet-planet gravitational forces are negligible compared with the Sun's pull — the Sun's mass dwarfs the planets'.
2. Helium: the electron-electron Coulomb force is comparable to the nucleus-electron force — the charges (e vs 2e) and distances are of the same order.
3. No small parameter exists to justify ignoring the e-e interaction → the one-electron machinery cannot be patched; a genuine many-body theory (quantum mechanics) is required.
4. Takeaway: NCERT Points to Ponder 4, dressed as a comparison — a favourite assertion-reason setup.