Ask Question
10 February, 13:46

In a simple model of the hydrogen atom, the electron moves in a circular orbit of radius 0.053nm around a stationary proton. part a how many revolutions per second does the electron make? hint: what must be true for a force that causes circular motion? express your answer using two significant figures.

+3
Answers (1)
  1. 10 February, 13:49
    0
    Here the force that is applied between the electron and proton is centripetal, so equate the two forces to determine the velocity.

    We know charge of the electron which for both Q1 and Q2, e = 1.60 x 10^-19 C

    The Coulombs Constant k = 9.0 x 10^9

    Radius r = 0.053 x 10^-9m = 5.3 x 10^-11 m

    Mass of the Electron = 9.11 x 10^-31

    F = k x Q1 x Q2 / r^2 = m x v^2 / r (centripetal force)

    ke^2 / r^2 = m x v^2 / r = > v^2 = ke^2 / m x r

    v^2 = ((1.60 x 10^-19) ^2 x 9.0 x 10^9) / (9.11 x 10^-31 x 5.3 x 10^-11)

    v^2 = 4.77 x 10^12 = 2.18 x 10^6 m/s

    Since one orbit is the distance,

    one orbit = circumference = 2 x pi x r; distance s = v x t.

    v x t = 2 x pi x r = > t = (2 x 3.14 x 5.3 x 10^-11) / (2.18 x 10^6)

    t = 33.3 x 10^-11 / 2.18 x 10^6 = 15.27 x 10^-17 s

    Revolutions per sec = 1 / t = 1 / 15.27 x 10^-17 = 6.54 x 10^15 Hz
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “In a simple model of the hydrogen atom, the electron moves in a circular orbit of radius 0.053nm around a stationary proton. part a how ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers