how many revolution per second does the electron make/?

Question

In the Bohr model of the hydrogen atom, an electron (m=9.1*10^-31kg orbits a proton at a distance of 5.3*10^-11m. The proton pulls on the electron with an electric force of 8.2*10^-8N.

Answer 1

In the Bohr model the electron is treated as though it were a classical particle, orbiting the proton due to the centripetal Coulomb force. The force of gravity is so small that it can be ignored.

The Coulomb force between two charges is k q1 q2 / r^2

where k is the Coulomb constant, q1, q2 are the charges of the electron and proton and r is the radius of the orbit. Solve for F.

Now Newton's Second Law in rotational form is F = m v^2 / r. You have F; m is the mass of the electron. Solve for v.

The circumference of the orbit is 2 pi r. You now have a distance per orbit and a velocity and can solve for time per orbit using

v t = d

That's the time per orbit; the reciprical would be the number of orbits per second. So the answer you want is 1 / t.

Yeah, it's a lot of steps but hopefully it makes sense.

Answer 2

The Angular momentum of the electron in Bohr's hydrogen atom is a constant which is equal to Planck's constant.
So the electron orbits the nucleus at an average velocity of
6.18 x10^6 meters per seconds. At a radius vector of one half angstrom its obital period is 1.51 x10^-16 seconds.
the average number of revolution of the electron per seconds is 6.58 x10^15 hertz. As the Electron changes mass it also change ts orbital radius vector thus its period and frequency also change accordingly.

The Electrostatic Gravity Power that pushes the Proton and Electron into a relative orbit is equal to the product of PLanck's constant and the electron's frequency of revolution square.

Answer 3

10 to the power of 34