An electron is released from rest at a distance of 9.00 cm from a fixed proton. How fast will the electron be moving when it is 3.00 cm from the proton? (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C², e = 1.60 × 10^-19 C, m_electron = 9.11 × 10^-31 kg, m_proton = 1.67 × 10^-27 kg)

Question

Quizplus · Accepted Answer

Using conservation of energy:
Initial potential energy = Final kinetic energy + Final potential energy
$\frac{k_e e^2}{r_i} = \frac{1}{2} m_e v_f^2 + \frac{k_e e^2}{r_f}$
where $r_i$ = initial distance = 9.00 cm
      $r_f$ = final distance = 3.00 cm
Solving for $v_f$, we get:
$v_f =$ sqrt$\frac{2k_ee^2}{m_e}( \frac{1}{r_f}-\frac{1}{r_i}) \approx 106$ m/s (rounded to 3 significant figures)

An Electron Is Released from Rest at a Distance of 9.00