Internal magnetic field experienced by H atom Electron

[Bosley]

Homework Statement

A 21 cm spectral line corresponds to the flipping of the electron in a hydrogen atom from having its spin parallel to the spin of the proton to having it anti-parallel. Find the internal magnetic field experienced by the electron in the hydrogen atom.

Homework Equations

lambda = 21 cm
f=c/lambda = 1.429 x 10^9 Hz
DeltaE = h f = 9.4686 x 10^-25 J

The Attempt at a Solution

According to Arthur Beiser's 1969 quantum book, the energy of an electron in a given quantum state will be higher or lower by V_m = (e*hbar/(2m))*B than its energy in the absence of the spin-orbit coupling.

I think that the DeltaE I found above would be equal to 2*V_m. Which would give me B = DeltaE*m/(e*hbar). But plugging in the numbers to that equation I get that B = .051 T.

This contradicts what Beiser says shortly thereafter, which is that the ground state hydrogen atom should have B approximately 14 T. He calculates this using B = mu_o*f*e/(2r) with f = 7 x 10¹⁵ where f is the number of times the hydrogen atom "sees" itself circled by the proton each second, and r = 5 x 10^-11 m. I don't know why my calculation contradicts this one, and if that contradiction is ok or if I have done something wrong.

Any insight would be appreciated.

 

[fzero]

The spin-orbit coupling is not responsible for the 21 cm line. The spin-orbit coupling refers to the coupling of the electrons intrinsic spin to its orbital angular momentum. Only states with nonzero orbital angular momentum are split by the spin-orbit interaction. This is one contribution to what's called fine structure of atom energy levels.

The 21 cm line is due to a splitting of the 1s state due to the coupling between the electron spin with the proton spin. The 1s state has no orbital angular momentum and there is no spin-orbit splitting for this state. The spin-spin coupling is much smaller than typical spin-orbit couplings and causes what is called hyperfine structure. I haven't checked your numbers, but it's natural to find a nuclear magnetic field that is much smaller than that due to angular momentum of the electron.