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#atomic structure

59 public questions tagged with this topic.

De Broglie Hypothesis and Quantization in Bohr Model

In Rutherford’s scattering experiment, what does the small number of alpha-particles undergoing head-on collisions sugge

**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 rarity of head-on collisions (1 in 8000 rebounding back) suggests that the nucleus, containing most of the mass and charge, is very small compared to the atom. 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 Nucleus occupies a

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

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Hydrogen Atom Properties - Radius, Speed and Energy

What is the orbital period of an electron in the \( n = 4 \) orbit if \( v_1 = 2.2 \times 10^6 \, \text{m/s} \) and \( r

**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. v₄ = (2.2 × 10⁶/4) = 5.5 × 10⁵ m/s . r₄ = 16 × 5.3 × 10⁻¹¹ = 8.48 × 10⁻¹⁰ m . T = (2π r₄/v₄) = (2 × 3.14 × 8.48 × 10⁻¹⁰/5.5 × 10⁵) ≈ 9.68 × 10⁻¹⁵ 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

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

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Bohr Model Energy Levels and Hydrogen Spectrum

What is a limitation of Bohr’s model when applied to the hydrogen atom’s spectrum?

**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. Bohr’s model cannot explain the relative intensities of spectral lines, as it does not account for transition probabilities. 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,

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

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Bohr Model Energy Levels and Hydrogen Spectrum

In Rutherford’s scattering experiment, what causes the large deflection of some alpha-particles?

**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. Large deflections occur due to the strong repulsive force from the positively charged nucleus when alpha-particles approach it closely. 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

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Bohr Model Energy Levels and Hydrogen Spectrum

In Rutherford’s experiment, what does the scattering of alpha-particles at small angles indicate?

**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. Small-angle scattering occurs when alpha-particles pass far from the nucleus (large impact parameter), experiencing weak repulsion. 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.

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

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Bohr Model Energy Levels and Hydrogen Spectrum

What does the line spectrum of hydrogen indicate about its atomic structure?

**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. The line spectrum suggests that electrons occupy discrete energy levels, emitting photons of specific wavelengths when transitioning between them. 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 Discr

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

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Semiconductors, Types and Energy Bands

The number of outer electrons in a Si crystal with \( N \) atoms is:

**Semiconductor properties** distinguish from conductors and insulators by temperature dependence and doping response. At 0 K intrinsic acts as insulator, conductivity due to thermally generated electron-hole pairs, number of outer electrons 4 for Si/Ge forming covalent bonds, each atom shares electrons, crystal with N atoms has 4N valence electrons, 2N bonds. Each Si atom has 4 valence electrons (2s and 2p). For N atoms, the total number of outer electrons is 4N , contributing to the covalent bonding in the lattice. Substituting values gives 4N, which matches expected behaviour for this semic

Ref: NCERT > Physics Book > Electronic Devices > Semiconductors, Types and Energy Bands

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Production of EM Waves and Hertz Experiment

Why are X-rays used in crystallography to study atomic structures?

**Hertz experiment** used induction coil connected to two rods with gap, spark produced oscillating charge, emitted EM wave, received by loop with gap sparking when E induced, measured wavelength by standing wave, demonstrated EM wave properties, validating Maxwell. X-rays have wavelengths comparable to atomic spacings, allowing them to diffract off crystal lattices and produce interference patterns that reveal atomic arrangements. Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields Comparable wavelengths, illustrating EM wave transverse nature and

Ref: NCERT > Physics Book > Electromagnetic Waves > Production of EM Waves and Hertz Experiment

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CHEMISTRY

How many electrons are in the L shell of an atom with 10 electrons?

Given: How many electrons are in the L shell of an atom with 10 electrons? These values define the system as per NCERT data. Formula: The L shell (n = 2) has 2s² 2p⁶ = 2 + 6 = 8 electrons.. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: For 10 electrons (neon): 1s² 2s² 2p⁶. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

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CHEMISTRY

What is the mass number of an ion with 13 protons, 14 neutrons, and a -1 charge?

Given: What is the mass number of an ion with 13 protons, 14 neutrons, and a -1 charge? Formula: Mass number (A) = protons + neutrons = 13 + 14 = 27. Substitution & Calculation: Charge affects electrons (14), not mass number. Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII), Chapter: Electrochemistry (Latest NCERT 2026-27), Topic: Faraday's laws, charge required to reduce Al³⁺, 3F = 3 × 96500 C

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CHEMISTRY

Which orbital designation corresponds to quantum numbers n = 4 and l = 3?

Given: Which orbital designation corresponds to quantum numbers n = 4 and l = 3? These values define the system as per NCERT data. Formula: For n = 4 and l = 3, the subshell is 4f (l = 0: s, l = 1: p, l = 2: d, l = 3: f).. This is standard NCERT relation. Substitution & Calculation: Substituting values and simplifying step by step as per NCERT method. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: d and f Block Elements, Topic: Lanthanoid contraction due to poor shielding of 4f orbitals.

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