An electron is released from rest at a distance of 9.00 cm from a proton. If the proton is held in place, how fast will the electron be moving when it is 3.00 cm from the proton? (mel = 9.11 x 10^-31 kg, e = 1.60 × 10^-19 C, k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²)

Question

Quizplus · Accepted Answer

We can use conservation of energy to solve this problem. At a distance of 9.00 cm, the electron has only potential energy, given by:
U = kqQ/r
where q is the charge of the electron (−e), Q is the charge of the proton (+e), r is the distance between them, and k is the Coulomb constant. Plugging in the numbers, we get:

U = (8.99 × 109 N∙m2/C2)(−1.60 × 10−19 C)(1.60 × 10−19 C)/(0.09 m) = −2.06 × 10−18 J

At a distance of 3.00 cm, the electron has both potential and kinetic energy. We can set the initial potential energy equal to the sum of the final potential and kinetic energies:

U = K + U'
where K is the final kinetic energy and U' is the final potential energy, at a distance of 3.00 cm. Plugging in the numbers, we get:

−2.06 × 10−18 J = (1/2)melv2 + (8.99 × 109 N∙m2/C2)(−1.60 × 10−19 C)(1.60 × 10−19 C)/(0.03 m)

Solving for v, we get:

v = 106 m/s

An Electron Is Released from Rest at a Distance of 9.00