#de Broglie

De Broglie Wavelength and Matter Waves

Which scientist proposed that moving particles of matter exhibit wave-like properties?

**De Broglie wavelength** λ = h/p = h/(m v) = h/√(2 m e V) for electron accelerated through V, for V=100 V λ= h/√(2 m e×100)=1.227/√V nm=0.1227 nm, for 50 V 0.173 nm, for particle mass 3.0×10⁻³⁰ kg v=10⁶ m/s λ=6.63×10⁻³⁴/(3×10⁻³⁰×10⁶)=2.21×10⁻¹⁰ m=0.221 nm, inversely proportional to momentum, property inversely proportional to de Broglie wavelength is momentum. Louis de Broglie proposed the wave-particle duality of matter, suggesting that particles like electrons have associated wavelengths ( λ = (h/p) ). Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ = h c/λ₀, λ = h/√(2

Ref: NCERT > Physics Book > Dual Nature of Matter > De Broglie Wavelength and Matter Waves

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Photon Energy, Momentum, Power and X-rays

A photon has a momentum of \( 1.8 \times 10^{-27} \, \text{kg m/s} \). What is its wavelength? (Take \( h = 6.63 \times

**X-rays from tube** maximum frequency f_max = e V/h, minimum wavelength λ_min = h c/(e V), Duane-Hunt law, for 10 kV f_max=1.6×10⁻¹⁹×10⁴/6.63×10⁻³⁴=2.41×10¹⁸ Hz, λ_min= c/f_max=0.124 nm, for 35 kV λ_min=1240/(35000) eV·nm? Actually 35 keV photon λ=1240/35000=0.0354 nm, for 15 kV 0.0827 nm, for 25 kV 0.0496 nm, for 30 kV 0.0413 nm. λ = (h/p) = (6.63 × 10⁻³⁴/1.8 × 10⁻²⁷) ≈ 3.683 × 10⁻⁷ m = 368.3 nm . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ = h c/λ₀, λ = h/√(2 m e V) and f_max = e V/h,

Ref: NCERT > Physics Book > Dual Nature of Matter > Photon Energy, Momentum, Power and X-rays

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Bohr Model Energy Levels and Hydrogen Spectrum

Which of the following statements is incorrect about de Broglie’s explanation of Bohr’s model?

**Bohr energy levels** E_n = -13.6/n² eV for hydrogen, negative indicating bound state, total energy = -13.6 eV ground state n=1, -3.4 eV n=2, -1.51 eV n=3, etc., photon energy for transition n_i → n_f is ΔE =13.6(1/n_f² -1/n_i²) eV, wavelength λ = hc/ΔE, h=6.6×10⁻³⁴ J·s, c=3×10⁸ m/s. Emission line spectrum characterized by discrete wavelengths because energy levels discrete. De Broglie’s hypothesis supports quantization by proposing standing waves, not continuous waves, which fit an integer number of wavelengths into the orbit. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, B

Ref: NCERT > Physics Book > Atoms and Nuclei > Bohr Model Energy Levels and Hydrogen Spectrum

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De Broglie Hypothesis and Quantization in Bohr Model

What does de Broglie’s hypothesis explain in the context of Bohr’s model?

**Bohr's quantization** angular momentum L = m v r = n h/2π, for n=2 L=2h/2π= h/π=2.11×10⁻³⁴ J·s, for n=5 L=5h/2π, de Broglie λ = h/p, p= m v, for first orbit v=2.2×10⁶ m/s, λ= h/(m v)=6.6×10⁻³⁴/(9.1×10⁻³¹×2.2×10⁶)=3.3×10⁻¹⁰ m, circumference 2πr=3.33×10⁻¹⁰ m, one wavelength fits for n=1. De Broglie’s hypothesis explains the quantization of angular momentum by proposing that electrons form standing waves, with the orbit circumference being an integral multiple of the wavelength. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields Qua

Ref: NCERT > Physics Book > Atoms and Nuclei > De Broglie Hypothesis and Quantization in Bohr Model

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