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De Broglie Hypothesis and Quantization in Bohr Model

This category gathers questions about the De Broglie hypothesis and its role in the Bohr model’s quantization rules. It covers how electron wavelength determines allowed orbital energies, linking wave‑particle duality to atomic structure. Ideal for students reviewing physics concepts.

30 questions

What is the volume of a nucleus with radius \( 4.8 \times 10^{-15} \, \text{m} \)? (Use \( \pi = 3.14 \))

**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. Volume = (4/3) π R³ . R³ = (4.8 × 10⁻¹⁵)³ = 1.105 × 10⁻⁴³ m³ . Volume = (4/3) × 3.14 × 1.105 × 10⁻⁴³ ≈ 4.63 × 10⁻⁴³ m³ . 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

What is the volume of a nucleus with radius \( 3.0 \times 10^{-15} \, \text{m} \)? (Use \( \pi = 3.14 \))

**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. Volume = (4/3) π R³ . R³ = (3.0 × 10⁻¹⁵)³ = 2.7 × 10⁻⁴⁴ m³ . Volume = (4/3) × 3.14 × 2.7 × 10⁻⁴⁴ ≈ 1.13 × 10⁻⁴³ m³ . 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

How much energy is released when \( 5 \, \text{g} \) of matter is converted into energy? (Given \( c = 3 \times 10^8 \,

**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. E = m c² . m = 5 × 10⁻³ kg , c² = 9 × 10¹⁶ m²/s² . E = 5 × 10⁻³ × 9 × 10¹⁶ = 4.5 × 10¹⁴ J . 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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What is the energy difference between the \( n = 4 \) and \( n = 2 \) states in a hydrogen atom? (Use \( E_n = -\frac{13

**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. E₄ = -0.85 eV , E₂ = -3.4 eV . Δ E = -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 model and nuclear binding energy systematics.

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

What does the absorption spectrum of a hydrogen atom reveal?

**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. The absorption spectrum shows dark lines at wavelengths where photons are absorbed, matching the emission line wavelengths, indicating specific energy level transitions. 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 Dark lines in a

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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

Which of the following statements is incorrect about Thomson’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. Thomson’s model does not include a nucleus; it proposes a uniform positive charge distribution, unlike Rutherford’s nuclear model. 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 Described as a plum pudding model, consistent with Bohr model and

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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 minimum energy required to excite a hydrogen atom from its ground state to the first excited state? (Use \(

**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. E₁ = -13.6 eV , E₂ = -3.4 eV . Δ E = -3.4 - (-13.6) = 10.2 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

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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.

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What is the angular momentum of an electron in the \( n = 3 \) state of a hydrogen atom? (Use \( h = 6.6 \times 10^{-34}

**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. L = n (h/2π) . For n = 3 : L = 3 × (6.6 × 10⁻³⁴/2 × 3.14) ≈ 3.15 × 10⁻³⁴ J·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 3.15 × 10⁻³⁴ J·s, consistent with

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

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