How Does the Zeeman Effect Alter the Wavelength of Hydrogen's Paschen Lines?

[fluidistic]

Homework Statement

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

Homework Equations

$\Delta E=m_l \mu _B B$.
$E=\frac{hc}{\lambda}$.

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 $m_l=1$.
So applying the first formula I gave, this gives $\Delta E \approx 1.85 \times 10^{-23}J=1.16\times 10^{-4}eV$.
Applying the second formula this gives me $\Delta \lambda \approx 0.01 m$.
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.

 

[fluidistic]

Nevermind, the approach is right. I just made a calculator mistake with the product "hc".

 

[Redbelly98]

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.

 

[fluidistic]

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.

 

[asteriatic]

First of all, great job trying to solve the problem and showing your work! It's important to always double check your calculations and make sure your results make sense. In this case, you are correct that the change in wavelength you calculated is too large. This is because you used the wrong value for the magnetic field.

The correct formula for the Zeeman effect is actually \Delta E=g\mu_B B, where g is the Lande g-factor. For the hydrogen atom, g=1, so your calculation for \Delta E should be 2 times smaller. Additionally, the value for the Bohr magneton \mu_B is 9.274\times 10^{-24}J/T, which is about 200 times smaller than the value you used. This means that your final result for \Delta \lambda should be about 400 times smaller, which is much more reasonable.

So, to summarize, your mistake was using the wrong values for g and \mu_B in your calculation. Keep up the good work and always double check your results!