Practice question
Question
What is the energy equivalent of \( 0.15 \, \text{g} \) of matter in Joules? (Given \( c = 3 \times
10^8 \, \text{m/s} \))
Explanation
**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. E = m c² . m = 0.15 × 10⁻³ kg = 1.5 × 10⁻⁴ kg , c² = 9 × 10¹⁶ m²/s² . E = 1.5 × 10⁻⁴ × 9 × 10¹⁶ = 1.35 × 10¹³ J . 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 1.35 × 10¹³ J,
Discussion
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