# Checking my reasoning on deriving fine sructure constant alpha

• Emspak
In summary: So in summary, the ratio of the electron velocity in the ground state of hydrogen atom to the speed of light is equal to the fine structure constant α, which can be expressed in terms of fundamental physical constants h, m, and c as α = e^2 / (hbar * c), where e is the electron charge and hbar is the reduced Planck's constant. This can be derived from Bohr's model of the hydrogen atom by solving for the velocity and substituting in the value for r.
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)

Emspak said:
$v = \frac{e^2 2\pi r}{n h}$ and we divide that by c to get α.
It won't be a fundamental physical constant if it depends on ##r## and ##n##.

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)

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

is that correct?
No. You made a mistake somewhere. Start from
Emspak said:
$v = \frac{nh}{2\pi m r}$
and substitute in the value of ##r## you got.

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).

Emspak said:
$$\frac{v}{c} = \frac{2 \pi e^2}{n h c} = \frac{e^2}{\hbar c} = \alpha$$
Correct, although you should get rid of the ##n## before getting to this point (or at least mention why it disappeared here).

Emspak said:
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!

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

## 1. What is the fine structure constant alpha?

The fine structure constant alpha, also known as the fine-structure constant or Sommerfeld's constant, is a dimensionless physical constant that characterizes the strength of the electromagnetic interaction between elementary particles. It is approximately equal to 1/137.

## 2. How is the fine structure constant alpha derived?

The value of alpha can be derived from several fundamental physical constants, including the speed of light, the Planck constant, and the elementary charge. The most commonly used method is through quantum electrodynamics (QED), which describes the behavior of electromagnetic interactions at the quantum level.

## 3. Why is the fine structure constant alpha important?

The value of alpha is crucial in understanding the properties of atoms, molecules, and other fundamental particles. It also plays a significant role in the development of theories in physics, such as the Standard Model, which explains the behavior of subatomic particles and their interactions.

## 4. Can the fine structure constant alpha change over time?

According to current theories, the value of alpha is considered to be constant and does not vary over time or space. However, some theories suggest that it may have changed in the early universe or in extreme conditions, such as near black holes.

## 5. How is the fine structure constant alpha measured?

The most accurate method of measuring alpha is through precision experiments using quantum electrodynamics, such as the quantum Hall effect or the anomalous magnetic moment of the electron. These experiments can determine the value of alpha with an uncertainty of less than one part per billion.

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