#hydrogen

De Broglie Hypothesis and Quantization in Bohr Model

A hydrogen atom emits a photon when an electron drops from \( n = 5 \) to \( n = 2 \). What is the energy of the photon?

**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. E₅ = -(13.6/5²) = -0.544 eV , E₂ = -3.4 eV . Δ E = -0.544 - (-3.4) = 2.856 eV ≈ 2.86 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 2.86 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

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De Broglie Hypothesis and Quantization in Bohr Model

An electron in a hydrogen atom falls from \( n = 4 \) to \( n = 2 \). What is the energy of the emitted photon? (Use \(

**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. E₄ = -(13.6/4²) = -0.85 eV , E₂ = -3.4 eV . Δ E = E₄ - E₂ = -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

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

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Atomic Models - Rutherford, Thomson and Bohr

According to Bohr’s model, what happens when an electron in a hydrogen atom transitions from a higher orbit to a lower o

**Thomson's plum pudding model** positive charge uniformly distributed in sphere with electrons embedded, positive charge spread, fails to explain large angle scattering observed. Bohr's model introduces stationary orbits with quantized angular momentum L = n h/2π, physical basis de Broglie standing wave condition 2πr = n λ, circumference fits n wavelengths, explains line spectrum. Bohr’s third postulate states that a photon is emitted with energy equal to the difference between the initial and final energy states. 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 > Atomic Models - Rutherford, Thomson and Bohr

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Partial Pressures and Gas Mixtures

A gas mixture contains 6 g of hydrogen and 48 g of oxygen. What is the ratio of their partial pressures?

**Partial pressure concept** for mixture of non-reacting gases, total pressure sum of partials, each gas behaves independently, kinetic theory still holds with effective n = Σ n_i, mean free path depends on total n and cross-sections, mixture properties weighted by mole tions. P = (μ RT)/(V), P_H₂P_O₂ = μ_H₂μ_O₂.μ_H₂ = (6)/(2) = 3 mol, μ_O₂ = (48)/(32) = 1.5 mol.Ratio = (3)/(1.5) = 2:1. Substituting values gives 2:1, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Partial Pressures and Gas Mixtures

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Partial Pressures and Gas Mixtures

A gas mixture has equal numbers of hydrogen and oxygen molecules at 300 K. What is the ratio of their rms speeds? (Molec

**Gas mixtures** ideal gas law applies to each component, P_total = Σ n_i R T/V, partial pressure P_i = n_i R T/V, mole tion X_i = n_i/n_total, P_i = X_i P_total, enabling calculation of individual pressures from composition, important for kinetic theory and chemistry. v_rms ∝ (1)/(√(m)), v_H₂v_O₂ = √(m_O)₂m_H₂ = √((32)/(2)) = √(16) = 4. Substituting values gives 4:1, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Partial Pressures and Gas Mixtures

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RMS Speed and Temperature Dependence

At what temperature is the rms speed of hydrogen molecules 2000 m/s? (Molecular mass of H₂ = 2 u, k_B = 1.38 × 10⁻²³ J K

**RMS speed** v_rms = √(3 R T/M) = √(3 k_B T/m) where M molar mass (kg/mol), m molecular mass (kg), k_B=1.38×10⁻²/³ J/K, R=8.314 J/mol·K, T absolute temperature (K). Proportional to √T and 1/√M, lighter gases faster at same T, e.g., H₂ faster than O₂, temperature increase raises v_rms as √T. v_rms = √((3k_B T)/(m)), m = 2 × 10⁻³⁶.02 × 10²³ = 3.32 × 10⁻²⁷ kg.2000² = 3 × 1.38 × 10⁻²/³ × T3.32 × 10⁻²⁷, T = 4 × 10⁶ × 3.32 × 10⁻²⁷/⁴.14 × 10⁻²/³ ≈ 321 K. Substituting values gives 321 K, which matches expected kinetic theory result, confirming mean free path

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > RMS Speed and Temperature Dependence

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Internal Energy of Ideal Gases

A gas mixture contains 16 g of oxygen and 4 g of hydrogen. What is the ratio of their partial pressures?

**Internal energy of ideal gas** U = f/2 n R T depends only on temperature, f degrees of freedom, n moles, R=8.314 J/mol·K, for monatomic f=3 U=3/2 n R T, diatomic f=5 at moderate T U=5/2 n R T, independent of pressure or volume, only T matters for ideal gas. P = (μ RT)/(V), P_O₂P_H₂ = μ_O₂μ_H₂.μ_O₂ = (16)/(32) = 0.5 mol, μ_H₂ = (4)/(2) = 2 mol.Ratio = (0.5)/(2) = 1:4. Substituting values gives 1:4, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Internal Energy of Ideal Gases

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Degrees of Freedom and Molar Specific Heat

What is the average translational kinetic energy of a hydrogen molecule at 700 K? (k_B = 1.38 × 10⁻²³ J K⁻¹)

**Molar specific heat** from equipartition C_v = f/2 R, C_p = f/2 R + R, γ = C_p/C_v =1+2/f, for f=3 γ=1.67, f=5 γ=1.4, f=6 γ=1.33, explaining specific heat variation with molecular structure, degrees of freedom determine heat capacity. Average translational KE = (3)/(2) k_B T.(3)/(2) × 1.38 × 10⁻²/³ × 700 = 1.449 × 10⁻²⁰ J. Substituting values gives 1.449 × 10⁻²⁰ J, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Degrees of Freedom and Molar Specific Heat

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Degrees of Freedom and Molar Specific Heat

A gas mixture has equal masses of hydrogen and argon at 300 K. What is the ratio of their rms speeds? (Molecular mass: H

**Molar specific heat** from equipartition C_v = f/2 R, C_p = f/2 R + R, γ = C_p/C_v =1+2/f, for f=3 γ=1.67, f=5 γ=1.4, f=6 γ=1.33, explaining specific heat variation with molecular structure, degrees of freedom determine heat capacity. v_rms ∝ (1)/(√(m)), v_H₂v_Ar = √(m_Ar)m_H₂.v_H₂v_Ar = √((39.9)/(2)) ≈ √(19.95) ≈ 4.47. Substituting values gives 4.47:1, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Degrees of Freedom and Molar Specific Heat

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Degrees of Freedom and Molar Specific Heat

A gas mixture contains 2 g of hydrogen and 32 g of oxygen. What is the ratio of their partial pressures?

**Specific heat relation** C_p - C_v = R for ideal gas per mole, Mayer's relation, due to work done at constant pressure, degrees of freedom include translational, rotational, vibrational, each quadratic term contributes ½ R to C_v. P = (μ RT)/(V), P_H₂P_O₂ = μ_H₂μ_O₂.μ_H₂ = (2)/(2) = 1 mol, μ_O₂ = (32)/(32) = 1 mol.Ratio = (1)/(1) = 1:1. Substituting values gives 1:1, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Degrees of Freedom and Molar Specific Heat

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Degrees of Freedom and Molar Specific Heat

The rms speed of oxygen molecules at a certain temperature is 500 m/s. What is the rms speed of hydrogen molecules at th

**Specific heat relation** C_p - C_v = R for ideal gas per mole, Mayer's relation, due to work done at constant pressure, degrees of freedom include translational, rotational, vibrational, each quadratic term contributes ½ R to C_v. v_rms = √((3k_B T)/(m)), so v_rms ∝ (1)/(√(m)).Ratio: v_rms,H₂v_rms,O₂ = √(m_O₂)m_H₂ = √((32)/(2)) = √(16) = 4.v_rms,H₂ = 4 × 500 = 2000 m/s . Substituting values gives 2000 m/s, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Degrees of Freedom and Molar Specific Heat

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CHEMISTRY

Which reaction involves the displacement of hydrogen?

In Ca + 2Hâ‚‚O -> Ca(OH)2 + Hâ‚‚, Ca displaces H (+1 to 0), a redox displacement reaction. This follows from NCERT principle where the relation explains the outcome clearly for students in simple steps.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

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