Question

In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm. (a) Find the magnitude of the electrostatic force of attraction, Fe between the electron and the proton. (b) Find the magnitude of the gravitational force of attraction Fg , between the electron and the proton, and find the ratio, Fe /Fg . me = 9.11 x 10-31kg, e = 1.602 x 10-19C mp = 1.67 x 10-27kg k = 9 x 109 Nm2 /C2 G = 6.67 x 10-11 Nm2 /kg2

Answer #1

in the bhor model of the hydrogen atom the electron is assumed
to orbit the proton in a circle at an average distance of
5.3x10^-11 m. the centripetal force keeping the electron in orbit
is due to the Coulomb force law. e= 1.6x10^-19C, electron = 9.11 x
10^-31
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the proton?
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An electron (mass m1 = 9.11 x 10-31 kg) and a proton (mass m2 =
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atom is placed in a vacuum chamber that is filled with a uniform
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The electron in a hydrogen atom can be thought of as orbiting
around the nucleus (a single proton) in a circular orbit of radius
5.28×10−11m with a velocity of constant magnitude
2.190×106m/s . At t= 0s a hydrogen
atom is placed in a vacuum chamber that is filled with a uniform
magnetic field of magnitude 3.3×1011T pointing in the
positive z -direction. The hydrogen atom is oriented such
that its electron is orbiting counter-clockwise in the
xy -plane at t= 0s .
At t= 12s ,...

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