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#alpha particle

6 public questions tagged with this topic.

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

In an alpha-particle scattering experiment, a 5.5 MeV alpha-particle approaches a gold nucleus (Z = 79). What is the app

**Rutherford's nuclear model** atom has small massive positively charged nucleus with electrons orbiting, size ratio atomic to nuclear ~10⁵, nucleus ~10⁻¹⁵ m atom ~10⁻¹⁰ m, most alpha particles with large impact parameter pass undeflected, small fraction >90° scatter from close approach, centripetal force provided by Coulomb attraction k Z e²/r², fails to explain stability because accelerating charge should radiate and collapse. Energy conservation: K = (2Ze²/4πepsilon₀ d) . d = (2Ze²/4πepsilon₀ K) . K = 5.5 MeV = 5.5 × 1.6 × 10⁻¹³ = 8.8 × 10⁻¹³ J . d = (2 × 79 × (1.6 × 10⁻¹⁹)² × 9 × 10⁹/8.8 ×

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

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

An alpha-particle with 4.0 MeV kinetic energy approaches a gold nucleus (Z = 79). What is the distance of closest approa

**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. d = (2Ze²/4πepsilon₀ K) . K = 4.0 × 1.6 × 10⁻¹³ = 6.4 × 10⁻¹³ J . d = (2 × 79 × (1.6 × 10⁻¹⁹)² × 9 × 10⁹/6.4 × 10⁻¹³) . d = (3.641 × 10⁻²⁸/6.4 × 10⁻¹³) ≈ 5.69 × 10⁻¹⁴ m ≈ 57 fm . 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 > Hydrogen Atom Properties - Radius, Speed and Energy