#angular momentum

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

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

What is the angular momentum of an electron in the \( n = 5 \) state of a hydrogen atom? (Use \( h = 6.6 \times 10^{-34}

**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. L = n (h/2π) . For n = 5 : L = 5 × (6.6 × 10⁻³⁴/2 × 3.14) ≈ 5.25 × 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 5.25 × 10⁻³⁴ J·s, consistent with

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

What is the angular momentum of an electron in the \( n = 2 \) 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 = 2 : L = 2 × (6.6 × 10⁻³⁴/2 × 3.14) ≈ 2.1 × 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 2.1 × 10⁻³⁴ J·s, consistent with

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

In Bohr’s model, what is the physical basis for the quantization of angular momentum?

**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 second postulate states that the angular momentum of the electron is an integral multiple of h/2π , introducing quantization to ensure stable orbits. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A

Ref: NCERT > Physics Book > Atoms and Nuclei > Atomic Models - Rutherford, Thomson and Bohr

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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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PHYSICS

A 4 kg mass rotates in a circle of radius 0.25 m with a linear speed of 2 m/s . What is its angular momentum about the n

Given: A 4 kg mass rotates in a circle of radius 0.25 m with a linear speed of 2 m/s . What is its angular momentum about the nter? These values define the system as per NCERT data. Formula: L = m v r. This is standard NCERT relation. Substitution & Calculation: m = 4 kg, v = 2 m/s, r = 0.25 m . L = 4 × 2 × 0.25 = 2 kg m²/s . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: System of Particles and Rotational Motion, Topic: Angular momentum L = m v r, rotating mass and moment of momentum. The section explains definitions, governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, units and illustrative examples.

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PHYSICS

What is the effect of increasing the mass of a rotating body on its angular momentum, if angular velocity remains consta

Angular momentum L = I omega, and I increases with mass. If omega is constant, L increases proportionally with mass. This follows from latest NCERT 2026-27 principle explaining the concept clearly for NEET students in simple steps as per rationalized syllabus.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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Moment of Inertia and Radius of Gyration

A 6kg particle moves with velocity v\=3j^m/s at r\=−4i^m. What is the magnitude of its angular momentum about the origin

L = r×p = |i^j^k^−400030| = k^((−4)×3−0×0) = −12k^kg m2/s. Magnitude = 12kg m2/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 12 kg m²/s. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.