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#nuclear volume

3 public questions tagged with this topic.

Hydrogen Atom Properties - Radius, Speed and Energy

What is the volume of a nucleus with radius \( 3.6 \times 10^{-15} \, \text{m} \)? (Use \( \pi = 3.14 \))

**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. Volume = (4/3) π R³ . R³ = (3.6 × 10⁻¹⁵)³ = 4.6656 × 10⁻⁴⁴ m³ . Volume = (4/3) × 3.14 × 4.6656 × 10⁻⁴⁴ ≈ 1.95 × 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 yields 1.95 × 10⁻⁴³ m³, 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 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

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 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

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

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