Zeeman effect exercise

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fluidistic
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Homework Statement


The [itex]\alpha[/itex] lines of Paschen in the hydrogen spectrum are due to transitions [itex]n=4 \to n=3[/itex]. Identify the allowed [itex]4p \to 3d[/itex] transitions and determine the change in wavelength for each transition if there's an external B field of 2T.


Homework Equations


[itex]\Delta E=m_l \mu _B B[/itex].
[itex]E=\frac{hc}{\lambda}[/itex].

The Attempt at a Solution


I graphed all transitions possible (it's an enormous mess).
Now say I want to calculate the difference of wavelength of with and without the magnetic field for the transition 4p, m=0 and 3d, m=1 (it's allowed). I have that [itex]m_l=1[/itex].
So applying the first formula I gave, this gives [itex]\Delta E \approx 1.85 \times 10^{-23}J=1.16\times 10^{-4}eV[/itex].
Applying the second formula this gives me [itex]\Delta \lambda \approx 0.01 m[/itex].
I know this result is totally senseless. It's way too big, enormous. From memory Paschen lines are in the near infrared so about 800 nm and a bit up. Nothing like 0.01m!
I really don't know what I'm doing wrong.
I would appreciate some help.
 

Answers and Replies

  • #2
fluidistic
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Nevermind, the approach is right. I just made a calculator mistake with the product "hc".
 
  • #3
Redbelly98
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Hi! Glad it worked out.

By the way, both eV energy units and the product hc comes up so frequently in quantum mechanics that it's a good idea to remember (or right down, wherever you have a list of physics constants) it's value:

hc = 1240. eV-nm​

The energy-wavelength conversions will go quicker that way.
 
  • #4
fluidistic
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Hi! Glad it worked out.

By the way, both eV energy units and the product hc comes up so frequently in quantum mechanics that it's a good idea to remember (or right down, wherever you have a list of physics constants) it's value:

hc = 1240. eV-nm​

The energy-wavelength conversions will go quicker that way.
Thank you for your concern. And yes, I agree with you, I'll get "hc" in eV*nm into my formula sheet.
 

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