Skip to content

#Bohr model

71 public questions tagged with this topic.

In the Bohr model, how many de Broglie wavelengths fit into the circumference of the \( n = 5 \) orbit?

**De Broglie hypothesis** λ = h/p, p=mv momentum, suggests electron as wave, in Bohr model circumference 2πr = n λ, standing wave condition, n wavelengths fit into orbit, for n=6, 6 wavelengths, for n=4, 4 wavelengths, explains quantization of angular momentum L = r p = r h/λ = r h n/(2πr)= n h/2π = n ħ, physical basis for Bohr quantization. 2π r_n = nλ . For n = 5 , number of wavelengths = 5. 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

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

In the Bohr model, what is the ratio of the kinetic energy of an electron in the \( n = 2 \) state to that in the \( n =

**Hydrogen atom radius** r_n = n² a₀, a₀=5.3×10⁻¹¹ m first Bohr radius, r₂=4a₀=2.12×10⁻¹⁰ m, ratio r₄/r₂ =16/4=4, r₃=9a₀, circumference 2πr_n =2π n² a₀, for n=3 circumference=2π×9×5.3×10⁻¹¹=3×10⁻⁹ m. Orbital period T =2πr/v, v_n = v₁/n, v₁=2.2×10⁶ m/s, T₂=2πr₂/v₂, v₂=1.1×10⁶ m/s, T₂≈1.21×10⁻¹⁵ s. K = (e²/8πepsilon₀ r) , r_n ∝ n² . K_n ∝ (1/n²) . Ratio = (K₂/K₁) = (1/2²/1/1²) = (1/4) . 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 0.25, consistent with Bohr model and nuclear binding energy systematics.

Ref: NCERT > Physics Book > Atoms and Nuclei > Hydrogen Atom Properties - Radius, Speed and Energy

What is the speed of an electron in the \( n = 5 \) orbit of a hydrogen atom if its speed in \( n = 1 \) is \( 2.2 \time

**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. v_n = (v₁/n) . For n = 5 : v₅ = (2.2 × 10⁶/5) = 4.4 × 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 MeV, evaluation yields 4.4 × 10⁵ m/s, consistent with Bohr model and nuclear binding energy systematics.

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

What is the frequency of revolution of an electron in the first orbit of a hydrogen atom if its speed is \( 2.2 \times 1

**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. v = (v/2π r) . v = (2.2 × 10⁶/2 × 3.14 × 5.3 × 10⁻¹¹) ≈ 6.6 × 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 6.6 × 10¹⁵ Hz, consistent with Bohr model and nuclear binding energy systematics.

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

What is the speed of an electron in the \( n = 3 \) orbit of a hydrogen atom if its speed in \( n = 1 \) is \( 2.2 \time

**De Broglie hypothesis** λ = h/p, p=mv momentum, suggests electron as wave, in Bohr model circumference 2πr = n λ, standing wave condition, n wavelengths fit into orbit, for n=6, 6 wavelengths, for n=4, 4 wavelengths, explains quantization of angular momentum L = r p = r h/λ = r h n/(2πr)= n h/2π = n ħ, physical basis for Bohr quantization. v_n = (v₁/n) . For n = 3 : v₃ = (2.2 × 10⁶/3) ≈ 7.33 × 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

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

What limitation of Bohr’s model prevents it from explaining the spectrum of helium?

**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. Bohr’s model applies only to hydrogenic (single-electron) atoms and cannot account for electron-electron interactions in multi-electron atoms like helium. 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 Inability to handle multi-electron atoms,

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

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

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

Ref: NCERT > Physics Book > Atoms and Nuclei > 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

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

Ref: NCERT > Physics Book > Atoms and Nuclei > 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 =

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

In the Bohr model, how many de Broglie wavelengths fit into the circumference of the \( n = 3 \) orbit?

**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. 2π r_n = nλ . For n = 3 , number of wavelengths = 3. 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 3, consistent with Bohr model and nuclear binding energy systematics.

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