Energy of different energy levels of hydrogen atom

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SUMMARY

The discussion centers on determining the energy levels of a hydrogen atom in the 2p state under a magnetic field of 0.10 T. It concludes that there are six distinct energy levels, derived from the magnetic quantum number (ml) values of -1, 0, and 1, each corresponding to two spin states (ms) of +/- 1/2. The energy levels are calculated using the formula U = -gmsμzB, where g is approximately -2 for electrons, and the initial energy E0 is -13.6/n^2, with n being 2. The results indicate that the energies are extremely small and negative, raising questions about the validity of the calculations.

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leroyjenkens
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Homework Statement


A hydrogen atom is in 2p state, how many different energy levels are there if a magnetic field of 0.10 T is applied to it, and what are their energies? Make sure spin is included.



Homework Equations


U = -gmsμzB

g is the Lande-g factor, which is about -2 for electrons
ms is spin, which is +/- 1/2
μz is -μbm, which is the Bohr magneton times mass.
Bohr magneton is equal to about 5.78x10-5
The mass is, I assume, the mass of the electron, which is about 9.109x10-31

And the initial energy of the ground state is E0 is -13.6/n^2

n is 2.

The Attempt at a Solution


I get 6 energy levels, because ml has -1,0,1 and each of those has a spin of +/- 1/2.

So what I did was just make the equation E0 + U, which gives me the initial energy, plus the energy that U provides. But I'm getting extremely small numbers for U. For example, for ml = 1, and the +1/2 spin, I'm getting, for U, 5.27x10-36. Added to E0, that's miniscule. Plus I'm always going to have a negative answer, since E0 is negative.

So I don't know why it's negative, and I'm not sure if my answers are correct, because they seem extremely small.

Thanks.
 
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leroyjenkens said:
I get 6 energy levels, because ml has -1,0,1 and each of those has a spin of +/- 1/2.
Because of spin-orbit coupling, ##m_l## and ##m_s## are not good quantum numbers. You need to look at the total angular momentum of the electron ##j## and its projection ##m_j##.
 
DrClaude said:
Because of spin-orbit coupling, ##m_l## and ##m_s## are not good quantum numbers. You need to look at the total angular momentum of the electron ##j## and its projection ##m_j##.

In what way are they not good quantum numbers?

For total angular momentum of the electron, do I use \frac{\sqrt{3}}{2}\hbar

Thanks.
 

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