Fine Structure Constant

[Jay-qu]

I posted this in an old thread and I think it went under the radar.. so Im reposting it here because I found it interesting :)

 

I was learning some atomic physics the other day and came across the fine structure constant and the equation for the velocity of an atomic electron. It goes like this:

 

[math]\frac{v}{c} = \frac{aZ}{n}[/math]

 

ok so that doesnt mean much like that, but if you sub in the fine structure constant of 1/137 and take the first orbital n=1...

 

[math]\frac{v}{c} = \frac{Z}{137}[/math]

 

so for this element we have:

 

[math]\frac{v}{c} = \frac{118}{137}=86%[/math]

 

:eek: thats damned fast!

 

But tell me, what happens in the n=1 orbital when (and if) we ever reach an atom with Z=137?

 

Will it turn out that the formula is only an approximation for small values of Z?

[Erasmus00]

To answer your question, lets look at where the formula comes from. The first thing we need to know is "how much energy does an electron in a hydrogenlike atom carry?"

 

This question can be succesfully answered with quantum mechanics. We need to solve Schroedingers equation for an electron in a 1/r potential.

 

[math] \frac{-\hbar}{2m}\nabla^2 \psi -\frac{Ze^2}{r}\psi = E\psi [/math]

 

Its important to realize is that this is where we are making out assumption that this is not a relativistic system. Schroedingers equation is a quantization of a non-relativistic form of the energy. The first term in the equation corresponds to 1/2 p^2/2m type kinetic energy terms, the second is just the potential and the right hand side is the energy.

 

We could solve this, but I think its interesting to note there is a semi-classical "rule of thumb" method of calculating that gets the right answer(the bohr model of the atom). Lets consider our system as an electron orbitting in a circle around a nucleus. In this case (classically) the electric force needs to provide the centripital acceleration of the electron, so F=ma becomes

 

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

 

Now we need to eliminate r, which can do by invoking a tiny bit of quantum mechanics. We use the de Broglie relation for the wavelength

 

[math] \lambda = \frac{h}{p} = \frac{h}{mv} [/math]

 

Notice that we have used the non-relativistic form of the momentum, hence our equations will not be good for v close to c.

 

Now, the circumference of the circle the particle is traveling in must be an integer multiple of the wavelength, or else the wave wouldn't "close" on itself when it traveled all the way around, so

 

[math] \lambda = \frac{2\pi r}{n} [/math]

 

which implies

 

[math] r = \frac {n}{\hbar n}{mv} [/math]

 

So using this equation plus our earlier force equation

 

[math]\frac{Ze^2 mv}{\hbar n} = mv^2 [/math]

[math]\frac{Ze^2}{\hbar n} = v [/math]

 

Dividing by c we get

 

[math]\frac{v}{c} = \frac{Z\alpha}{n} [/math]

 

Where alpha is the fine structure constant. The key thing I've tried to emphasize is that you have to invoke non-relativistic energies to get this result, no matter how you try to calculate this. One way to try and correct this result would be to take the relatvistic kinetic enegy and expand to first order in (V^2/c^2) and try to solve the schroedinger equation perturbatively.

-Will

[Jay-qu]

Thanks Erasmus, a very satisfying answer - I can take that back to my lecturer now :)