Ignoring spin, what is its orbital magnetic moment of an electron in a hydrogen atom in the 2p orbital?

I know that a magnetic dipole moment is given by $$\mu=\frac e{2m}I$$ and that the angular momentum is $$\frac {m_jh}{2\pi}.$$ However, I have also seen that angular momentum $I$ is given by $$I=\frac h{2\pi}\sqrt{l(l+1)}.$$

What's going on here? Are both expressions for angular momentum right? Which one do I use to answer the question?

(This question was asked in a problem sheet that I have been asked to do over the vacation. See below for context of question). Your expression "the angular momentum is $m_j \hbar$" (where $\hbar = h/2 \pi$) is incorrect. This quantity is the projection of the angular momentum on the $z$-axis; it represents the direction of the spin. This is why it corresponds to the $m$ quantum number, not the $\ell$ quantum number in the $| n \ell m \rangle$ basis.
$$\mathbf{\mu} = \frac{-e}{2 m} \mathbf{L}$$ Here $\mathbf{L}$ is the total angular momentum operator. Its full expression is a vector, but the eigenvectors of its square are $\hat{L}^2 | n \ell m \rangle = \hbar^2 \ell( \ell + 1) | n \ell m \rangle$. However, it's very possible that the problem you're doing wants some sort of energy due to the interaction of the dipole with a magnetic field, which is given by $\mathbf{ \mu \cdot B}$. If the magnetic field is taken to be in the $z$-direction, as it often is, this reduces to: $$E =\frac{-e}{2m} B L_z$$ Now what you want is indeed the projection on the $z$-axis, and the operator $L_z$ does have the eigenvalues you mention, $\hat{L}_z | n \ell m \rangle = m \hbar | n \ell m \rangle$.
• Yes, you would have to use the overall operator $\mathbf{J} = \mathbf{L} + \mathbf{S}$ and its projection $m_J$. Apr 3 '15 at 20:09
• Do you want to find $\mathbf{J}^2$ or $\mathbf{J}_z$? To get $\mathbf{J}_z$ is easy since $m_J = \mathbf{J} \cdot z = (\mathbf{L} + \mathbf{S}) \cdot z = m_\ell + m_s$. Apr 3 '15 at 20:12