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.