Effective Magnetic Field in Ground-State Hydrogen Atom?

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Homework Statement



A hydrogen atom in its ground state actually has two possible, closely spaced energy levels because
the electron is in the magnetic field B of the proton (the nucleus). Accordingly, an energy is associated
with the orientation of the electron's magnetic moment (μ) relative to B, and the electron is said to be
either spin up (higher energy) or spin down (lower energy) in that field. If the electron is excited to
the higher-energy level, it can de-excite by spin-flipping and emitting a photon. The wavelength
associated with that photon is 21 cm. (Such a process occurs extensively in the Milky Way galaxy,
and reception of the 21 cm radiation by radio telescopes reveals where hydrogen gas lies between
stars.) What is the effective magnitude of B as experienced by the electron in the ground-state
hydrogen atom?


Homework Equations



E=Bμ
B(total)= B(int) + B(ext)
E=hf

The Attempt at a Solution


I determined the energy with the wavelength given. However, I do not know how to tackle the effective magnitude of B .. The groundstate of an hydrogen atom => 13.6 eV
How can I determine μ? Or do I not need it?
 
on Phys.org
Of course, you need the magnetic moment of the electron. There's a caveat when doing this. Look at a textbook on atomic physics under the key work "Thomas precession". It's a huge relativistic effect (factor of 2) which you wouldn't expect when thinking within naive non-relativistic QT!