Zeeman Effect: Splitting of Lyman-α Wavelength

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The discussion focuses on calculating the splitting of the Lyman-α line (121.6 nm) under a magnetic field of 1 Tesla using the Zeeman effect. The user applied the derivative method to derive the change in wavelength, resulting in a final answer of 121.6 ± 0.0007 nm. There was a debate over the correctness of the exponent used for the Bohr magneton and the necessity of squaring the wavelength in calculations. Participants suggested additional resources for understanding the Zeeman effect in hydrogen and discussed the implications of quantum states and spectral lines resulting from the splitting. The conversation emphasizes the complexity of the topic and the importance of accurate calculations.
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Homework Statement


The Lyman-α line (n = 2 → n = 1) has a wavelength of 121.6nm in the absence of a magnetic field.
When B = 1 Tesla, into how many wavelengths will this split and what are their wavelengths?

(You may need the values
μ B = 9.274 × 10 −24 J/T, h̄ = 1.054 × 10 −34 Js and c = 2.998 × 10 8 m/s.)

Homework Equations


##\Delta E=\mu _B\cdot B\cdot m##

The Attempt at a Solution


Have I done this correctly in my lecture note they use Taylor series which I understand but I have use the derivative method as follows:

$$E=\frac{hc}{\lambda }$$

$$\Delta E=\frac{hc\Delta \lambda }{\lambda ^2}$$

$$\Delta \:\lambda =\:\frac{\mu _B\cdot B\cdot m\cdot \lambda ^2}{hc}$$

subbing in the values

$$\Delta \:\lambda =\:\frac{\left(9.274\cdot 10^{-24}\right)\left(1\right)\left(1\right)\left(121.6\cdot 10^{-9}\right)^2}{\left(6.626\cdot 10^{-34}\right)\left(3\cdot 10^8\right)}=6.89\cdot 10^{-13}m$$

so then my final answer is

$$\lambda \pm \Delta\lambda=121.6 \pm 0.0007 nm$$

is there an advantage to using the Taylor series approach?
 
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Why did you switch the exponent in ## \mu_B ## from ## 10^{-24} ## to ## 10^{-27} ##? ## \\ ## I think ## 10^{-24} ## is correct, and you introduced a ## 10^{-3} ## error. ## \\ ## And your Taylor series does work very well. (It actually carries the name that you chose on PF). :)
 
Charles Link said:
Why did you switch the exponent in ## \mu_B ## from ## 10^{-24} ## to ## 10^{-27} ##? ## \\ ## I think ## 10^{-24} ## is correct, and you introduced a ## 10^{-3} ## error. ## \\ ## And your Taylor series does work very well. (It actually carries the name that you chose on PF). :)

Sorry I have edited it, when I do latex quickly I tend to mistake thing sorry.
 
And you need to square ## \lambda ##. Clearly a "typo".
 
I have squared ##\lambda## have I not? is what I have done method wise correct?
 
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Taylor_1989 said:
I have squared ##\lambda## have I not? is what I have done method wise correct?
I checked the arithmetic quickly=I get about ## .7 \cdot 10^{-12}## m=.0007 nm . I'll recheck the arithmetic, but that's what I got. (I'm working without a calculator). ## \\ ## Edit: Your answer looks correct ! :) ## \\ ## Additional item: You get ## m_l=\pm 1, n=2 ## going to ## m_l=0, n=1 ##, accounting for the splitting. I think you can also get ## m_l=0, n=2 ##, going to ## m_l=0, n=1 ##. How many spectral lines does that give? One question that would take additional research is if there is any ## m_s ## splitting. It could be worthwhile to google the Zeeman effect in hydrogen. ## \\ ## And you are on the right track, but it appears this one gets slightly complicated. See https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Quantum_Mechanics/Zeeman_Effect/5:_Zeeman_Effect_in_Hydrogen_atom (Some of the detail is hard to read in this "link"). This one is ## n=3 ## to ## n=2 ##, but may also be helpful. https://www.scribd.com/document/262518364/Zeeman-Effect-in-Hydrogen ## \\ ## See also: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/zeeman.html ## \\ ## And here's what may be the best, and most applicable of the bunch: https://en.wikipedia.org/wiki/Zeeman_effect#Example:_Lyman_alpha_transition_in_hydrogen It looks like it gets a little complicated sorting out the different allowed transitions, as well as perhaps determining since you have a relatively strong magnetic field, that some of the fine structure that occurs might be considered to be the same spectral line, even though there may be additional fine splittings.
 
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Charles Link said:
And you need to square ## \lambda ##. Clearly a "typo".

Charles Link said:
I checked the arithmetic quickly=I get about ## .7 \cdot 10^{-12}## m=.0007 nm . I'll recheck the arithmetic, but that's what I got. (I'm working without a calculator). ## \\ ## Edit: Your answer looks correct ! :) ## \\ ## Additional item: You get ## m_l=\pm 1, n=2 ## going to ## m_l=0, n=1 ##, accounting for the splitting. I think you can also get ## m_l=0, n=2 ##, going to ## m_l=0, n=1 ##. How many spectral lines does that give? One question that would take additional research is if there is any ## m_s ## splitting. It could be worthwhile to google the Zeeman effect in hydrogen. ## \\ ## And you are on the right track, but it appears this one gets slightly complicated. See https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Quantum_Mechanics/Zeeman_Effect/5:_Zeeman_Effect_in_Hydrogen_atom (Some of the detail is hard to read in this "link"). This one is ## n=3 ## to ## n=2 ##, but may also be helpful. https://www.scribd.com/document/262518364/Zeeman-Effect-in-Hydrogen ## \\ ## See also: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/zeeman.html ## \\ ## And here's what may be the best, and most applicable of the bunch: https://en.wikipedia.org/wiki/Zeeman_effect#Example:_Lyman_alpha_transition_in_hydrogen It looks like it gets a little complicated sorting out the different allowed transitions, as well as perhaps determining since you have a relatively strong magnetic field, that some of the fine structure that occurs might be considered to be the same spectral line, even though there may be additional fine splittings.

Thank you very much, and I will start reading the information you link now, much apprecatied :)
 
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