Skip to content

#orbit circumference

2 public questions tagged with this topic.

The radius of the first orbit in a hydrogen atom is \( 5.3 \times 10^{-11} \, \text{m} \). What is the circumference of

**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. r_n = n² r₁ , r₄ = 4² × 5.3 × 10⁻¹¹ = 8.48 × 10⁻¹⁰ m . Circumference = 2π r₄ = 2 × 3.14 × 8.48 × 10⁻¹⁰ ≈ 5.33 × 10⁻⁹ 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

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

In a hydrogen atom, the radius of the first orbit is \( 5.3 \times 10^{-11} \, \text{m} \). What is the circumference of

**Energy in hydrogen** total E_n = -13.6/n² eV, kinetic K = +13.6/n² eV, potential U = -27.2/n² eV, ratio K/U = -1/2, total negative indicates bound, zero at ionization, negative total means electron bound, requires energy to free. For n=4 E=-0.85 eV, U=-1.7 eV, K=0.85 eV, potential energy twice total negative. r_n = n² r₁ , r₃ = 3² × 5.3 × 10⁻¹¹ = 4.77 × 10⁻¹⁰ m . Circumference = 2π r₃ = 2 × 3.14 × 4.77 × 10⁻¹⁰ ≈ 3.0 × 10⁻⁹ m . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE

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