આ MCQ મોડ્યુલ આના પર આધારિત છે: De Broglie Explanation
De Broglie Explanation
આ મૂલ્યાંકન આના પર આધારિત હશે: De Broglie Explanation
મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.
De Broglie Explanation
12.15 The Mystery of Bohr's Second Postulate
Bohr's three postulates work miraculously well for hydrogen. But where does the curious quantisation rule \(L = nh/(2\pi)\) come from? Bohr himself had no answer — it was an inspired guess. Why, of all possible numbers, must angular momentum equal an integer multiple of \(h/2\pi\)?
The answer came in 1924 from Louis de Broglie, while writing his PhD thesis. If electrons have a wave nature (λ = h/p), then a stable orbit is one in which the electron-wave forms a standing wave on the circle. Like a vibrating violin string clamped at both ends, only certain wavelengths are allowed.
12.16 de Broglie's Standing-Wave Argument
Imagine the electron's matter-wave wrapped around a circular orbit of radius \(r_n\). For the wave to interfere constructively with itself after one trip around — that is, for the orbit to be stable — the circumference must contain a whole number of wavelengths:
\[2\pi r_n = n\lambda \quad (n = 1, 2, 3, \ldots)\]Substituting de Broglie's wavelength \(\lambda = h/(m_e v)\):
\[2\pi r_n = \frac{nh}{m_e v}\]Rearranging:
This is exactly Bohr's quantisation condition! The mysterious "integer angular momentum" is just a fancy way of saying "the matter wave fits the circle smoothly." The mystery is gone; quantum is geometry.
A non-allowed (forbidden) orbit
If the circumference \(2\pi r\) is not a whole number of wavelengths, the wave returns out of phase after each circuit. Successive trips would cancel each other by destructive interference. Over time the wave dies out — the orbit cannot be stable. Only the allowed integer orbits survive, exactly recovering Bohr's quantisation.
12.17 Limitations of Bohr's Model
Bohr's model was a triumph for hydrogen. But its successes ended there. Detailed observations soon revealed cracks:
(a) Limited to one-electron systems
Bohr's formula correctly predicts the energies of hydrogen and hydrogen-like ions (He⁺, Li²⁺, Be³⁺) — any system with one electron orbiting Z protons. For neutral helium (2 electrons), the model fails utterly. Inter-electron repulsion is not included; a complete description requires the many-body Schrödinger equation.
(b) Cannot explain spectral line intensities
Bohr's model says which lines exist, but not how bright each one is. In reality, some Balmer lines are extremely intense, others nearly invisible. Predicting these intensities requires transition probabilities (matrix elements of the dipole moment) — an output of full quantum mechanics, not Bohr's simple postulates.
(c) Fine structure
Examined under high resolution, every "single" Bohr line splits into two or more closely-spaced lines — the so-called fine structure. This splitting (~10⁻⁴ eV) is due to the electron's spin coupling to its orbital motion (spin-orbit interaction) and to relativistic corrections, neither of which appears in Bohr's model.
(d) Zeeman and Stark effects
Place a hydrogen discharge tube in a strong magnetic field: each Bohr line splits into several components — the Zeeman effect. A strong electric field similarly splits lines (Stark effect). Bohr's model has nothing to say about either, since it allows only one possible orbit shape (a circle) at each n.
(e) Semi-classical, not consistently quantum
Bohr postulates that orbits don't radiate (a quantum statement) but otherwise treats the electron as a tiny billiard ball moving on a definite circular path (a classical statement). The two pictures don't sit easily together. The complete quantum-mechanical description, due to Erwin Schrödinger (1926), abandons the idea of definite orbits altogether. Instead, it describes the electron by a wavefunction ψ(r) whose squared modulus |ψ|² gives the probability of finding it at point r. The "orbits" of Bohr are replaced by orbitals — three-dimensional probability clouds.
| Phenomenon | Bohr (1913) | Schrödinger (1926) |
|---|---|---|
| Hydrogen spectrum | Predicted | Predicted |
| Ionisation energy 13.6 eV | Predicted | Predicted (with same numerical value) |
| Helium spectrum | Fails | Predicted |
| Spectral line intensities | Cannot | Predicted via matrix elements |
| Fine structure | Cannot | Predicted (with Dirac equation) |
| Zeeman / Stark splitting | Cannot | Predicted |
| Electron position | Definite circular orbit | Probability cloud (orbital) |
Verify de Broglie's statement: in the n-th Bohr orbit of hydrogen, the circumference is exactly n times the de Broglie wavelength of the orbiting electron.
r₃ = 9 × 0.529 = 4.761 Å. v₃ = 2.19×10⁶/3 = 7.30×10⁵ m/s.
λ = h/(mₑv) = 6.626×10⁻³⁴/(9.11×10⁻³¹ × 7.30×10⁵) = 9.97×10⁻¹⁰ m.
2πr₃ = 2π × 4.761×10⁻¹⁰ = 2.992×10⁻⁹ m.
2πr₃ / λ = 2.992×10⁻⁹ / 9.97×10⁻¹⁰ = 3.00 ✓ — exactly n wavelengths fit!
Interactive — Standing Matter Waves on a Circle
Slide the quantum number n. The animation draws \(n\) full wavelengths around the orbit; observe how, for non-integer values (off the click-stops), the wave is discontinuous and would cancel itself.
Worked Examples
Calculate the de Broglie wavelength of a hydrogen electron in the n=2 orbit and verify that exactly two wavelengths fit around the orbit.
λ = 6.626×10⁻³⁴/(9.11×10⁻³¹ × 1.095×10⁶) = 6.65×10⁻¹⁰ m.
r₂ = 4 × 0.529 = 2.116 Å. 2πr₂ = 1.330×10⁻⁹ m.
2πr₂/λ = 1.330×10⁻⁹/6.65×10⁻¹⁰ = 2.00 ✓.
The first ionisation energy of helium (He → He⁺ + e⁻) is 24.6 eV — substantially less than the 54.4 eV that Bohr's formula predicts using Z = 2. Why?
The Balmer-α line is observed under high resolution to consist of (at least) two components separated by Δλ ≈ 0.014 nm. Estimate the energy splitting in eV.
For Hα: λ = 656 nm, E = 1.89 eV.
ΔE = 1.89 × 0.014/656 ≈ 4.0×10⁻⁵ eV ≈ 40 μeV. This tiny gap is the spin-orbit splitting Bohr cannot explain.
Competency-Based Questions
Q1. de Broglie's standing-wave condition gives:
Q2. Bohr's model is unable to explain:
Q3. (Short Answer) Explain in one sentence why de Broglie's idea makes Bohr's quantisation natural.
Q4. (Fill in the blank) The fine-structure splitting of spectral lines arises because the electron has ______ in addition to orbital motion.
Q5. (HOT) For an electron in the n=4 orbit, how many de Broglie wavelengths fit around the orbit, and what is the corresponding orbital angular momentum (in units of ℏ)?
Assertion–Reason Questions
Options: (A) Both true, R correct explanation of A. (B) Both true, R not the correct explanation. (C) A true, R false. (D) A false, R true.
Assertion: Bohr's quantisation rule arises from a standing-wave condition for the electron's matter wave.
Reason: The de Broglie wavelength is λ = h/(mv).
Assertion: Bohr's model accurately predicts the spectrum of helium.
Reason: Helium has two electrons.
Assertion: Schrödinger's wave-mechanical description does away with Bohr's idea of definite circular orbits.
Reason: The electron is described by a wavefunction whose square gives the probability density of finding it at a point.
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Physics — CBSE Class XII Sample Paper 1 (2025-26)
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