Hydrogen atom: How fast must a hydrogen atom be traveling for its kinetic energy to be just enough to excite the ground-state atom to its first excited state in a collision? (1 eV = 1.60 × 10^-19 J, m_h ≈ m_proton = 1.67 × 10^-27 kg)

Question

Quizplus · Accepted Answer

The energy difference between the ground state and the first excited state of a hydrogen atom is approximately 10.2 eV. Converting this to joules gives 10.2 eV × 1.60 × 10^-19 J/eV = 1.63 × 10^-18 J. Using the kinetic energy formula KE = 0.5 * m * v^2, and solving for v, we get v = sqrt((2 * KE) / m) = sqrt((2 * 1.63 × 10^-18 J) / (1.67 × 10^-27 kg)) ≈ 44.2 km/s.

Hydrogen Atom: How Fast Must a Hydrogen Atom Be Traveling