#quantum number

Hydrogen Atom Properties - Radius, Speed and Energy

What is the kinetic energy of an electron in the \( n = 4 \) state of a hydrogen atom? (Use \( E_n = -\frac{13.6}{n^2} \

**Angular momentum** L_n = n h/2π, h=6.6×10⁻³⁴ J·s, L₂=2×6.6×10⁻³⁴/2π=2.11×10⁻³⁴ J·s, speed v_n = e²/(2 ε₀ h) ×1/n ≈2.2×10⁶/n m/s, kinetic energy ½ m v² =13.6/n² eV, illustrating Bohr model predictions for hydrogen-like atoms. E₄ = -0.85 eV , K = -E₄ = 0.85 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 0.85 eV, consistent with Bohr model and nuclear binding energy systematics.

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

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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.

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

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De Broglie Hypothesis and Quantization in Bohr Model

An electron in a hydrogen atom has an angular momentum of \( 1.05 \times 10^{-34} \, \text{J·s} \). What is its principa

**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. L = n (h/2π) . 1.05 × 10⁻³⁴ = n × (6.6 × 10⁻³⁴/2 × 3.14) . n = 1 . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and

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

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Quantum Numbers - Principal Azimuthal Magnetic Spin

Which quantum number determines the shape of an orbital?

The azimuthal quantum number l determines the shape of an orbital.

Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Quantum Numbers - Principal Azimuthal Magnetic Spin

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