De Broglie Hypothesis and Quantization in Bohr Model

De Broglie Hypothesis and Quantization in Bohr Model

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

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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

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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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De Broglie Hypothesis and Quantization in Bohr Model

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.

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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 continuous spectrum, consistent with Bohr

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De Broglie Hypothesis and Quantization in Bohr Model

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

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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 nuclear binding energy systematics.

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De Broglie Hypothesis and Quantization in Bohr Model

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.

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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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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.

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De Broglie Hypothesis and Quantization in Bohr Model

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

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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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