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#Coulomb barrier

3 public questions tagged with this topic.

Hydrogen Atom Properties - Radius, Speed and Energy

What is the key requirement for nuclei to undergo fusion?

**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. Fusion requires nuclei to overcome the Coulomb barrier (electrostatic repulsion between positively charged nuclei), which is achieved by providing high kinetic energy through elevated temperatures. 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 > Hydrogen Atom Properties - Radius, Speed and Energy

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Hydrogen Atom Properties - Radius, Speed and Energy

The kinetic energy of an alpha-particle is 7.7 MeV. If it approaches a nucleus with atomic number 79, what is its distan

**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. d = (2Ze²/4πepsilon₀ K) . K = 7.7 × 1.6 × 10⁻¹³ = 1.232 × 10⁻¹² J . d = (2 × 79 × (1.6 × 10⁻¹⁹)² × 9 × 10⁹/1.232 × 10⁻¹²) . d = (3.641 × 10⁻²⁸/1.232 × 10⁻¹²) ≈ 2.95 × 10⁻¹⁴ m ≈ 30 fm . Using E_n = -13.6/n²

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

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Hydrogen Atom Properties - Radius, Speed and Energy

An alpha-particle with kinetic energy 6.0 MeV approaches a gold nucleus (Z = 79). What is the distance of closest approa

**Hydrogen atom radius** r_n = n² a₀, a₀=5.3×10⁻¹¹ m first Bohr radius, r₂=4a₀=2.12×10⁻¹⁰ m, ratio r₄/r₂ =16/4=4, r₃=9a₀, circumference 2πr_n =2π n² a₀, for n=3 circumference=2π×9×5.3×10⁻¹¹=3×10⁻⁹ m. Orbital period T =2πr/v, v_n = v₁/n, v₁=2.2×10⁶ m/s, T₂=2πr₂/v₂, v₂=1.1×10⁶ m/s, T₂≈1.21×10⁻¹⁵ s. d = (2Ze²/4πepsilon₀ K) . K = 6.0 × 1.6 × 10⁻¹³ = 9.6 × 10⁻¹³ J . d = (2 × 79 × (1.6 × 10⁻¹⁹)² × 9 × 10⁹/9.6 × 10⁻¹³) . d = (3.641 × 10⁻²⁸/9.6 × 10⁻¹³) ≈ 3.79 × 10⁻¹⁴ m ≈ 38 fm . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R

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

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