- #1

Emspak

- 243

- 1

## Homework Statement

From Bohr’s model of hydrogen atom, derive the ratio of the electron velocity in the ground state of hydrogen atom to the speed of light, which is the fine structure constant α, in terms of fundamental physical constants such as h, m, and c.

## Homework Equations

Kinetic energy in a Bohr atom: [itex]\frac{Ze^2}{r^2}[/itex]

kinetic energy of a body revolving around a center: [itex]\frac{mv^2}{r}[/itex]

momentum of a body revolving around a center: [itex]L=mvr[/itex]

## The Attempt at a Solution

So we set the kinetic energy equal to that of the Bohr atom (based on the Coulomb force)

[itex]\frac{mv^2}{r}=\frac{Ze^2}{r^2}[/itex]

Since we are dealing with hyrogen Z=1, so [itex]\frac{mv^2}{r}=\frac{e^2}{r^2}[/itex] and [itex]r=\frac{e^2}{mv^2}[/itex]

Bohr's model says the angular momentum around a hydrogen atom is [itex]\frac{nh}{2\pi}=n\hbar[/itex]

The total energy of the atom is [itex]T+U = E_T[/itex]

and angular momentum can be expressed: [itex]mvr = \frac{nh}{2\pi}[/itex]

so [itex]v = \frac{nh}{2\pi m r}[/itex] which we can substitute into the equation for r and get

[itex]r = \frac{e^2}{m(\frac{n h}{2\pi mr})^2}=\frac{4\pi^2 e^2 m r^2}{n^2 h^2}[/itex] which yields [itex]r = \frac{n^2 h^2}{4\pi^2 e^2 m}[/itex]

That's the Bohr radius. We can plug that into the original relation we had for kinetic energy.

[itex]\frac{mv^2}{r}=\frac{Ze^2}{r^2}[/itex]

which gets us

[itex]\frac{n^2 h^2}{4\pi^2 e^2 m r^2}=\frac{e^2}{mv^2}[/itex]

and moving it all around algebraically

[itex]v^2 = \frac{e^2 4\pi^2 e^2 m r^2}{n^2 h^2 m}[/itex]

[itex]v = \frac{e^2 2\pi r}{n h}[/itex] and we divide that by c to get α.

So my question is if I did this correctly. (I sense that someone will tell me I did something wrong, but I want to see if its conceptual or arithmetical mistake)