Bohr Model Energy Levels and Hydrogen Spectrum

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, BE = Δm c² and 1 u

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Bohr Model Energy Levels and Hydrogen Spectrum

The radius of the first orbit in a hydrogen atom is \( 5.3 \times 10^{-11} \, \text{m} \). What is the radius of the fif

**Excitation** energy required to go from n=1 to n=3 is 12.09 eV, from ground to n=∞ ionization 13.6 eV, state n=∞ means ionized, electron free with zero energy, highest level reached by electron beam energy determines which levels can be excited, e.g., 11 eV beam from ground can reach n=2 (10.2 eV) but not n=3 (12.09 eV), so max n=2. r_n = n² r₁ . For n = 5 : r₅ = 5² × 5.3 × 10⁻¹¹ = 25 × 5.3 × 10⁻¹¹ = 1.325 × 10⁻⁹ m . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀

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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

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Bohr Model Energy Levels and Hydrogen Spectrum

What is the speed of an electron in the \( n = 4 \) orbit of a hydrogen atom if its speed in \( n = 1 \) is \( 2.2 \time

**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. v_n = (v₁/n) . For n = 4 : v₄ = (2.2 × 10⁶/4) = 5.5 × 10⁵ m/s . 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

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Bohr Model Energy Levels and Hydrogen Spectrum

In the Bohr model, what is the ratio of the radius of the \( n = 5 \) orbit to the \( n = 1 \) orbit?

**Excitation** energy required to go from n=1 to n=3 is 12.09 eV, from ground to n=∞ ionization 13.6 eV, state n=∞ means ionized, electron free with zero energy, highest level reached by electron beam energy determines which levels can be excited, e.g., 11 eV beam from ground can reach n=2 (10.2 eV) but not n=3 (12.09 eV), so max n=2. r_n = n² r₁ . r₅ = 25 r₁ , r₁ = r₁ . Ratio = (r₅/r₁) = 25 . 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 =

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Bohr Model Energy Levels and Hydrogen Spectrum

An electron beam of 12.0 eV excites a hydrogen atom in the ground state. What is the highest energy level reached? (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. E₁ = -13.6 eV . E_n = -13.6 + 12.0 = -1.6 eV . -1.6 = -(13.6/n²) ⇒ n² ≈ 8.5 ⇒ n = 2 (since E₂ = -3.4 eV < -1.6 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

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Bohr Model Energy Levels and Hydrogen Spectrum

What is the role of the impact parameter in Rutherford’s alpha-particle scattering?

**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. The impact parameter determines the scattering angle; a smaller impact parameter leads to a larger deflection due to closer approach to the nucleus. 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

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Bohr Model Energy Levels and Hydrogen Spectrum

In Rutherford’s nuclear model, why do most alpha-particles pass through the gold foil without deflection?

**Excitation** energy required to go from n=1 to n=3 is 12.09 eV, from ground to n=∞ ionization 13.6 eV, state n=∞ means ionized, electron free with zero energy, highest level reached by electron beam energy determines which levels can be excited, e.g., 11 eV beam from ground can reach n=2 (10.2 eV) but not n=3 (12.09 eV), so max n=2. Most of the atom is empty space, with the nucleus occupying a very small volume, so most alpha-particles do not encounter the nucleus and pass through undeflected. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE

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Bohr Model Energy Levels and Hydrogen Spectrum

What assumption in Rutherford’s model contradicts classical electromagnetic theory?

**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. Rutherford assumes electrons orbit the nucleus like planets, but classical theory predicts they would radiate energy and collapse, not remain stable. 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 =

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Bohr Model Energy Levels and Hydrogen Spectrum

In Bohr’s model, what happens to the energy required to ionize a hydrogen atom as the electron’s orbit number increases?

**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. As n increases, the energy becomes less negative (closer to zero), so less energy is required to ionize the atom from higher orbits. 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

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Bohr Model Energy Levels and Hydrogen Spectrum

What is the frequency of a photon emitted when an electron drops from \( n = 3 \) to \( n = 1 \) in a hydrogen atom? (Us

**Excitation** energy required to go from n=1 to n=3 is 12.09 eV, from ground to n=∞ ionization 13.6 eV, state n=∞ means ionized, electron free with zero energy, highest level reached by electron beam energy determines which levels can be excited, e.g., 11 eV beam from ground can reach n=2 (10.2 eV) but not n=3 (12.09 eV), so max n=2. Δ E = 12.09 eV = 1.9344 × 10⁻¹⁸ J . nu = (Δ E/h) = (1.9344 × 10⁻¹⁸/6.6 × 10⁻³⁴) ≈ 2.93 × 10¹⁵ Hz . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE

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Bohr Model Energy Levels and Hydrogen Spectrum

Which of the following statements is correct about the absorption process in a hydrogen atom?

**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. Absorption occurs when an electron jumps to a higher energy level by absorbing a photon matching the energy difference between levels. 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 Electron jumps to a higher energy

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