#electron transition

De Broglie Hypothesis and Quantization in Bohr Model

A hydrogen atom emits a photon when an electron drops from \( n = 5 \) to \( n = 2 \). What is the energy of the photon?

**Quantization basis** de Broglie standing wave requires constructive interference, integer wavelengths in orbit, otherwise destructive, so only certain radii allowed r_n = n² a₀, a₀=0.53 Å, angular momentum L = n h/2π, de Broglie explains why orbits are stationary - electron wave closed on itself. E₅ = -(13.6/5²) = -0.544 eV , E₂ = -3.4 eV . Δ E = -0.544 - (-3.4) = 2.856 eV ≈ 2.86 eV . 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 2.86 eV, consistent with Bohr model and nuclear binding energy systematics.

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

View Details

De Broglie Hypothesis and Quantization in Bohr Model

An electron in a hydrogen atom falls from \( n = 4 \) to \( n = 2 \). What is the energy of the emitted photon? (Use \(

**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. E₄ = -(13.6/4²) = -0.85 eV , E₂ = -3.4 eV . Δ E = E₄ - E₂ = -0.85 - (-3.4) = 2.55 eV . 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 2.55 eV, consistent with Bohr

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

View Details

Bohr Model Energy Levels and Hydrogen Spectrum

What is the energy of a photon emitted when an electron drops from \( n = 5 \) to \( n = 4 \) in a hydrogen atom? (Use \

**Hydrogen transitions** example n=3→n=1 ΔE=13.6(1-1/9)=12.09 eV, photon 12.09 eV, λ=1240/12.09≈102.6 nm Lyman series, n=3→n=2 ΔE=1.89 eV Balmer visible Hα 656 nm. Absorption photon energy must match difference, if atom in n=2 absorbs 1.89 eV jumps to n=3, if absorbs 12.75 eV from ground 1→4 because -13.6+12.75=-0.85 eV = -13.6/16. E₅ = -0.544 eV , E₄ = -0.85 eV . Δ E = -0.544 - (-0.85) = 0.306 eV ≈ 0.31 eV . 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 0.31 eV, consistent with

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

View Details

Bohr Model Energy Levels and Hydrogen Spectrum

What is the frequency of a photon emitted when an electron drops from \( n = 2 \) to \( n = 1 \) in a hydrogen atom? (Us

**Hydrogen transitions** example n=3→n=1 ΔE=13.6(1-1/9)=12.09 eV, photon 12.09 eV, λ=1240/12.09≈102.6 nm Lyman series, n=3→n=2 ΔE=1.89 eV Balmer visible Hα 656 nm. Absorption photon energy must match difference, if atom in n=2 absorbs 1.89 eV jumps to n=3, if absorbs 12.75 eV from ground 1→4 because -13.6+12.75=-0.85 eV = -13.6/16. Δ E = 10.2 eV = 1.632 × 10⁻¹⁸ J . nu = (Δ E/h) = (1.632 × 10⁻¹⁸/6.6 × 10⁻³⁴) ≈ 2.47 × 10¹⁵ Hz . 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

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

View Details

Atomic Models - Rutherford, Thomson and Bohr

What is the wavelength of the photon emitted when an electron in a hydrogen atom drops from \( n = 3 \) to \( n = 2 \)?

**Rutherford's nuclear model** atom has small massive positively charged nucleus with electrons orbiting, size ratio atomic to nuclear ~10⁵, nucleus ~10⁻¹⁵ m atom ~10⁻¹⁰ m, most alpha particles with large impact parameter pass undeflected, small fraction >90° scatter from close approach, centripetal force provided by Coulomb attraction k Z e²/r², fails to explain stability because accelerating charge should radiate and collapse. E₃ = -1.51 eV , E₂ = -3.4 eV . Δ E = 1.89 eV = 1.89 × 1.6 × 10⁻¹⁹ = 3.024 × 10⁻¹⁹ J . λ = (hc/Δ E) = (6.6 × 10⁻³⁴ × 3 × 10⁸/3.024 × 10⁻¹⁹) ≈ 6.56 ×