# The radius of an electron orbit in helium

1. Jan 12, 2010

### Kennalj

can someone explain how bohr used the mass of the nucleus in helium to develop a ratio of 4.0016 of the original rydberg constant for hydrogen? I can't seem to find the proof anywhere, I read vaguely that he found this value by calculating the increased charge in the nucleus and using the mass of the nucleus in his calculation to say that the nucleaus and electron orbit eachother around a centre of mass, but whenever i try to do the calculations i keep getting a value of 4.002 instead of 4.0016

2. Jan 12, 2010

### ytuab

I'm a little in a hurry now, so I'm sorry if I misunderstand you.

You mean the helium ion (He+)?
In the helium ion, the relativistic effect is a little bigger than the hydrogen
(the relativistic energy change: He+ --- about 0.0026eV, H ---- 0.000....eV,
So the neutral He is between them.)

Did you consider this relativistic effect when calculating the energy?

3. Jan 12, 2010

### Kennalj

yea sorry I meant the helium ion. How would you go about applying relativity? Velocity is an unknown at this point isnt it?

4. Jan 13, 2010

### ytuab

In the Bohr-Sommerfeld model, the electron has the velocity like the classical particle.
But they are not the same things, because the electrons of Bohr model are influenced by the de Broglie's waves.

In the Bohr model, the relativistic energy of the hydrogen-like atom is,

$$W = m_{0}c^2(1/\sqrt{1-\frac{v^2}{c^2}} -1 ) - k\frac{Ze^2}{r}$$

The concrete calculation method is here (Wikipedia).

The result is,

$$W = m_{0}c^2(1 + \frac{\alpha^2 Z^2}{(n_{r}+\sqrt{n_{\phi}^2 - \alpha^2 Z^2})^2})^{-1/2} -m_{0}c^2$$

Accidentally this solution is completely the same as the solution of the Dirac equation of the hydrogen-like atoms.

In page 272 of the Principle of Quantum Mechanics by Dirac,
--------------------------------------------------------------
This formula gives the discrete energy-levels of the hydrogen spectrum and was first obtained by Sommerfeld working with Bohr's orbit theory.
--------------------------------------------------------------

If you compare this with the solution of the Schrodinger equation (or the usual Bohr model), the relativistic energy change (as I said in #2) is obtained.
In QM, the idea of the velocity is vague due to HUP. But the relativistic effect was actually seen. It's difficult to explain.

5. Jan 13, 2010

### Bob S

See page 254 in Dirac Principle of Quantum Mechanics (Oxford (Clarendon Press), 1930). It has only 257 pages. There are several later (four?) editions.
Bob S

6. Jan 13, 2010

### ytuab

Oh, yes. the book I have is the fourth edition (international series of monographs on physics).

In the section of "Relativistic theory of the electron".

(You have the 1st edition?)

7. Jan 13, 2010

### Bob S

Title page shows "1930"
Preface dated 29 May 1930 initialed PAMD
Inside front cover handwritten name & date Aug '30
Last page (#257) talks about "electron and proton disappearing simultaneously, their energy being emitted in the form of radiation."
Bob S

8. Jan 14, 2010

### ytuab

Bob, thanks. You have the 1st one.

Kennalj, I should have added more explanation (the original paper of Sommerfeld explained this in detail).

It is very interesting that the solution of the Bohr-Sommerfeld model which doesn't have the electron spin at all is completely consistent with the solution of the Dirac equation which contain the spin-orbital interaction.
Strange to say, there are many "accidental coincidences" in the development of QM.

The first condition is,

$$p_{\varphi}=mr^2 \dot{\varphi}, \quad m=\frac{m_{0}}{\sqrt{1-\beta^2}}, \quad \beta=\frac{v}{c}$$

Change the rectangular coodinates into the polar coordinates,

$$x = r cos \varphi, \qquad y = r sin \varphi$$

The nucleus is at the origin. The equation of motion is (the Coulomb force condition),

$$\frac{d}{dt}m\dot{x}= - \frac{kZe^2}{r^2}cos \varphi, \quad \frac{d}{dt}m\dot{y}= - \frac{kZe^2}{r^2}sin \varphi$$

Using the next condition (the angular momentum $$p_{\varphi}$$ is the constant),

$$\frac{d}{dt}= \frac{d\varphi}{dt} \frac{d}{d\varphi}= \frac{p_{\varphi}}{mr^2} \frac{d}{d\varphi}$$

So the equation of the motion is ($$u= 1/r$$),

$$\frac{d}{dt}m\dot{x}= - \frac{p_{\varphi}^2}{mr^2}(u+\frac{d^2 u}{d\varphi^2}) cos \varphi$$

In the case of y, change the upper cos into sin. Combine this with the Coulomb force condition,

$$\frac{d^2 u}{d \varphi^2}+u = \frac{kZe^2 m_{0}}{p_{\varphi}^2} \frac{1}{\sqrt{1-\beta^2}}$$

Using the energy $$W$$ (of #4) and erase the $$\beta$$, the solution can be expressed as,

$$u = \frac{1}{r} = C (1+ \epsilon cos \gamma \varphi)$$

And, the condition of the quantization is, (using the partial integration)

$$\oint p_{r}dr= p_{\varphi} \epsilon^2 \gamma \oint \frac{sin^2 \varphi d \varphi}{(1+\epsilon cos \varphi)^2} = p_{\varphi} \gamma \oint (\frac{1}{1+\epsilon cos \varphi}-1) d\varphi=n_{r} h$$

And, we should use the following mathematical formula, too,

$$\frac{1}{2\pi} \oint \frac{d \varphi}{1+ \epsilon cos \varphi} = \frac{1}{\sqrt{1-\epsilon^2}}$$

Combine all, the result of the energy W is obtained.

9. Jan 15, 2010

### Kennalj

$$p_{\varphi}=mr^2 \dot{\varphi}, \quad m=\frac{m_{0}}{\sqrt{1-\beta^2}}, \quad \beta=\frac{v}{c}$$

that first bit about momentum, where is that from?

10. Jan 16, 2010

### ytuab

Kennalji, the angular momentum is,

$$p_{\varphi} = p_{\perp} r = m v_{\perp} r = m(r \dot{\varphi})r = mr^2\dot{\varphi}$$

First, there is an Lorentz-invariant equation of motion which is valid irrespective of our observer's motions.

$$m_{0} \frac{d\omega_{\mu}(\tau)}{d\tau} = f_{\mu}(\tau) \qquad(1)$$

Here, $$\tau$$ is the Lorentz-invariant proper time (not t).
$$\omega_{\mu}, f_{\mu}$$ are the "four-velocity" and "four-vector force" which change as the four-vector under the Lorentz transformation.
These are, (i = x, y, or z)

$$\omega_{i}(\tau)=\frac{v_{i}(t)}{\sqrt{1-v^2/c^2}}, \quad f_{i}(\tau)=\frac{F_{i}}{\sqrt{1-v^2/c^2}}$$

Here, $$v_{i}$$ and $$F_{i}$$ are the velocity and force of Newtonian mechanics. And,

$$\frac{d\omega_{i}(\tau)}{d\tau}=\frac{d\omega_{i}(\tau(t))}{dt}/\frac{d\tau(t)}{dt}=\frac{1}{\sqrt{1-v^2/c^2}} \frac{d}{dt}(\frac{v_{i}(t)}{\sqrt{1-v^2/c^2}}){$$

Substituting this and $$f_{i}(\tau)$$ into the equation(1), we arrive at the following relation,

$$m_{0}\frac{d}{dt}(\frac{v_{i}(t)}{\sqrt{1-v^2/c^2}})= F_{i}$$

This means that we should use $$m_{0}\frac{v_{i}(t)}{\sqrt{1-v^2/c^2}}$$ instead of $$m_{0}v_{i}(t)$$ as the momentum(p) which we observe.

The Lorentz contraction of the changing radius(r) seems to occur. But the radius(r) belongs to our inertial system. Unless the radius(r) part of the air is separated from the surrounding air and moves toward us as a independent air stick, the contraction wouldn't occur.

And unless we ourselves move around the nucleus like the electron, we don't need to think about the gravity(which is too weak) and the inertia force like the centrifugal force.
So we can use the usual special relativity in the angular momentum as I said in #8.

Last edited: Jan 16, 2010
11. Apr 25, 2010

### sezw

hi i am a grade 12 student taking physics 30 (grd 12 level in alberta canada) can anyone give me e/m? the charge to mass ratio? today is april 25th and on the 27th i will be going to the university of calgary with my class to do the university labs one of them is charge to mass ratio of the electron and the other is a hydrogen balmerseiries lab involving spectrum lines. i am going over one of the pre lab booklets and question two asks me to calculate the accepted value for e/m there are several equations in the booklet but all involve coils or potential difference and i have been given no variables u guys seem to know your **** can u help?

12. Apr 25, 2010