Checking my reasoning on deriving fine sructure constant alpha

[Emspak]

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: $\frac{Ze^2}{r^2}$
kinetic energy of a body revolving around a center: $\frac{mv^2}{r}$
momentum of a body revolving around a center: $L=mvr$

The Attempt at a Solution

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

$\frac{mv^2}{r}=\frac{Ze^2}{r^2}$

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

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

The total energy of the atom is $T+U = E_T$

and angular momentum can be expressed: $mvr = \frac{nh}{2\pi}$
so $v = \frac{nh}{2\pi m r}$ which we can substitute into the equation for r and get

$r = \frac{e^2}{m(\frac{n h}{2\pi mr})^2}=\frac{4\pi^2 e^2 m r^2}{n^2 h^2}$ which yields $r = \frac{n^2 h^2}{4\pi^2 e^2 m}$

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

$\frac{mv^2}{r}=\frac{Ze^2}{r^2}$

which gets us

$\frac{n^2 h^2}{4\pi^2 e^2 m r^2}=\frac{e^2}{mv^2}$

and moving it all around algebraically

$v^2 = \frac{e^2 4\pi^2 e^2 m r^2}{n^2 h^2 m}$

$v = \frac{e^2 2\pi r}{n h}$ 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)

 

[DrClaude]

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

It won't be a fundamental physical constant if it depends on $r$ and $n$.

 

[Emspak]

OK, but since it is hydrogen (n=1) if I plug the expression for r into the one I got for v:

$$v=\frac{e^2 2 \pi}{n h} \frac{n^2 h^2}{4 \pi^2 e^2 m} = \frac{n h}{2 \pi m}$$

so that $$\alpha = \frac{n h}{2 \pi m c }$$ and since n=1

$$\alpha = \frac{h}{2 \pi m c } = \frac {\hbar}{mc}$$

is that correct? (I figure n had to be 1 since we are talking about hydrogen and a ground state)

 

[DrClaude]

$$\alpha = \frac{h}{2 \pi m c } = \frac {\hbar}{mc}$$

is that correct?

No. You made a mistake somewhere. Start from

$v = \frac{nh}{2\pi m r}$

and substitute in the value of $r$ you got.

 

[Emspak]

OK, doing that,

$v=\frac{nh}{2\pi m r}$

and

$r=\frac{n^2 h^2}{4\pi^2 e^2 m}$

So putting them together

$$v=\frac{nh}{2 \pi m } \frac{4 \pi^2 e^2 m}{n^2 h^2} = \frac{2 \pi e^2}{n h}$$

which would mean that

$$\frac{v}{c} = \frac{2 \pi e^2}{n h c} = \frac{e^2}{\hbar c} = \alpha$$

yes? assuming i got it righ this time I didn't need to do the second set of steps in my original derivation, I could have gone from when I got v originally rather than needing to go all the way to r.

(also, it's ok to assume n=1, right? We are talking about H).

 

[DrClaude]

$$\frac{v}{c} = \frac{2 \pi e^2}{n h c} = \frac{e^2}{\hbar c} = \alpha$$

Correct, although you should get rid of the $n$ before getting to this point (or at least mention why it disappeared here).

assuming i got it righ this time I didn't need to do the second set of steps in my original derivation, I could have gone from when I got v originally rather than needing to go all the way to r.

Indeed, I don't know why you did that! Your reasoning should go like this: You find an equation for $v$, which has a dependence on $r$. You then try to resolve this, and fortunately you have another equality which involves $r$. You solve that one, substitute $r$ in your equation for $v$, and see what you get. In this case, the job is then done!

 

[Emspak]

Yes, I think I got too hung up on needing the Bohr radius itself, which isn't necessary for this particular problem.