#nucleus size

Radioactive Decay, Nuclear Forces and Stability

What is the radius of a nucleus with mass number 16? (Given \( R_0 = 1.2 \times 10^{-15} \, \text{m} \))

**Nuclear force** strong, short-range ~1 fm, attractive, charge independent, saturated in large nuclei because each nucleon interacts only with neighbors, not all others, so BE/A saturates ~8 MeV, primary factor limiting stable nuclei size is Coulomb repulsion between protons growing as Z² vs strong force saturating, beyond Z≈83 no stable nuclei, competition between Coulomb and strong. The radius of a nucleus is given by R = R₀ A¹/³ , where A = 16 . A¹/³ = 16¹/³ = (2⁴)¹/³ = 2⁴/³ ≈ 2.52 . R = 1.2 × 10⁻¹⁵ × 2.52 ≈ 3.0 × 10⁻¹⁵ m . Using E_n = -13.6/n² eV, r_n = n²

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

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Nuclear Size, Density and Structure

Which factor primarily determines the nuclear radius?

**Nuclear density** nearly constant because R ∝ A^{1/3} so volume ∝ A, mass ∝ A, ratio constant, ~10¹⁷ kg/m³, 10¹⁴ times water density, shows nucleus compact, nuclear force short-range saturated, mass number 16 radius ~3×10⁻¹⁵ m, mass number from radius R=5.4×10⁻¹⁵ m => A=(R/R₀)³=(4.5)³=91. The nuclear radius is given by R = R₀ A¹/³ , where A (mass number) is the key factor determining the size, as the radius scales with the cube root of the number of nucleons. 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 > Nuclear Size, Density and Structure

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Nuclear Size, Density and Structure

What is the radius of a nucleus with mass number 200? (Given \( R_0 = 1.2 \times 10^{-15} \, \text{m} \))

**Nuclear size** order 10⁻¹⁵ m femtometer, atomic size 10⁻¹⁰ m, ratio 10⁵, nucleus contains protons and neutrons bound by strong force, density independent of A indicates incompressibility, R₀ determined from electron scattering experiments. R = R₀ A¹/³ . A = 200 , A¹/³ = 200¹/³ ≈ 5.85 . R = 1.2 × 10⁻¹⁵ × 5.85 ≈ 7.0 × 10⁻¹⁵ m . 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 7.0 × 10⁻¹⁵ m, consistent with Bohr model and nuclear binding energy systematics.

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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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De Broglie Hypothesis and Quantization in Bohr Model

In Rutherford’s model, what is the ratio of the size of the nucleus to the size of the atom if the nuclear radius is \(

**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. Ratio = (nuclear radius/atomic radius) = (10⁻¹⁵/10⁻¹⁰) = 10⁻⁵ . 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 10⁻⁵, consistent with Bohr

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