# Calculate the change in energy vs. B for each state separately?

1. Aug 16, 2008

Over a He-lamp a weak vertical magnetic field ( B = 300mT) is beeing applied. The light from the lamp is beeing studies with a high resolution spectrometer in the direction of the B-field. What will the wavelength difference between the observed Zeeman components be in the transmission 1s2p 1P - 1s3s 1S at 728.13nm

I realize that since we look in the direction of the B-field I understand that we'll only be able to see sigma-transmissions (delta M_j = -1 or +1). I've also calculated that g_j will be 1 for 1P and 3/2 for 1S.

But from there I dont really know how to continue.

2. Aug 16, 2008

### Redbelly98

Staff Emeritus
Re: Zeemancomponents

Can you calculate the change in energy vs. B for each state separately?

3. Aug 16, 2008

Re: Zeemancomponents

You tell me.
When I look in my formula sheet I find.

"For weak fields, hfs:
E_ZE = g_F * my_B * B * M_F"

I suppose that E_ZE is the energy separation? But I dont see how I can calculate g_F, since it contains the nuclear spin I, which I dont have. And I'm not even sure if thats the right way to go. I'm kind of calculating in the dark here.

4. Aug 16, 2008

### Redbelly98

Staff Emeritus
Re: Zeemancomponents

E_ZE is the shift in energy of a particular level.

Helium has zero nuclear spin, so there is no hfs. There is fine structure however. There should be a similar formula involving g_J and m_J.

5. Aug 17, 2008

Re: Zeemancomponents

Ok, this starts to make sence. I figured that S only have M_j = 0, which means that the're only 2 transmissions, so I calculated the difference to E_ZE(M_j = 1) - E_ZE(M_j = -1) and got delta f = 8.398GHz, which I've got confirmed is right.

but they ask for the wavelength difference, so there's probably some more work to do. If I didnt know that this gives me the wrong answear I would just go with lambda = c/(delta f). But that doesnt seem right. Why?

they say that they observe the transmission at 728.13nm, what does that really mean? The solution states the following.
lambda = c/f => delta lambda = (-) lambda^2/c * delta f.

First of all I dont understand why they do that delta lambda calculation, and when I try to get to the same conclusion I go like this
delta lambda = c/f_1 - c/f_2 = cf_2/(f_1 * f_2) - cf_1/(f_1*f_2) = c*delta f * 1/(f_1*f_2). And the only way I can connect this to the result they show in the solution is if c^2/(f_1*f_2) = lambda, but why would it? And why does the lambda from the text come in here?

A lot of blanks still remain.

6. Aug 17, 2008

### Redbelly98

Staff Emeritus
Re: Zeemancomponents

Good, that's reassuring.

Because lambda is really c/f, not c/(delta f). That is the basis for this calculation, along with:

$$\Delta \lambda = \lambda_1 - \lambda_2 = \frac{c}{f_1} -\frac{c}{f_2}$$

as you use in your derivation below.

The transition (not transmission) between the two states occurs at a wavelength of 728.13nm. Therefore the spectrum of light from the lamp contains this wavelength, and it is observed in the spectrometer.

So far so good ...

It's actually lambda^2, not lambda. It's an approximation.

Another useful piece of information is that the change in wavelength is a small fraction of the actual wavelength. Likewise, the change in frequency is a small fraction of the actual frequency. You can verify this be comparing your calculated frequency change of 8.398 GHz to the actual frequency of c/728.13nm.
What that does is make f1 and f2 nearly equal to each other and to c/lambda, where lambda is the value given in the problem statement. You can use that to get your expression in terms of lambda: