1. An electron in its ground state orbits the proton of a Hydrogen atom at an av
ID: 1440646 • Letter: 1
Question
1. An electron in its ground state orbits the proton of a Hydrogen atom at an average distance of 52.9 pm. (For this problem take the electron’s orbit around the proton to be a circle with a radius of 52.9 pm.) a) What is the ratio of the electric force between the electron and the proton to the gravitational force between the electron and the proton? (FE / Fg = ?) b) Does this ratio change if the distance between the electron and proton were to change? c) What is the orbital speed of the electron?
Explanation / Answer
A) FE/FG = (kQq/r2 )/(GMm/r2)
=kQq/GMm =
(9×109 ×1.6×10-19 ×1.6 ×10-19 )/( 6.6×10-11 ×1.6 ×10-27 ×9×10-31)
=2.4 × 1039
B)No...the ratio does not change as we can see that it does not depend on distance
C) since the ratio is huge...gravitational force is negligible
Only electrostatic force will be considered for circular motion
kQq/r2 = mv2/r
v2 = kQq/rm
= 9 ×10^9×1.6×10^-19×1.6×10^-19/(52.9×10^-12×9.1×10^-31)
=4.786 ×10^12
v= 2.188 × 106 m/s this is the orbital speed
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.