# Ignoring spin, consider an electron in a hydrogen 2p orbital, what is its orbital magnetic moment?

I know that a magnetic dipole moment is given by $$\mu=\frac {-e}{2m}I$$ and that the z component of angular momentum is $$m_j\hbar.$$ However, I have also seen that angular momentum $I$ is given by $$I=\hbar\sqrt{l(l+1)}.$$ Is this also the component in the z direction or is it different? 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).

The first equation you have is for spin. Let's look at it in a operator form

To properly treat the upcoming discussion, quantum electrodynamics is introduced to describe the eigenstates of spin. The eigenstates of spin has a direct correlation with spin. Formally we depict it as the following

$$μ_B =-g \frac {eℏ}{2m} \frac { \boldsymbol {S}}{ℏ}$$

where μ is the magnetic moment operator, g is the g factor, ℏ is plank’s constant over 2π, m is the mass of the particle and S is the spin operator.

For orbital magnetic moment (your question) you use:

$$\mu =\hbar\sqrt{l(l+1)} \boldsymbol {\mu_B}$$

Recall:

s: l=0 p: l=1 d: l=2 f: l=3

So plugging in:

$$\mu =\hbar\sqrt{1(1+1)} \boldsymbol {\mu_B}$$ $$\mu =\hbar\sqrt{2} \boldsymbol {\mu_B}$$

(Units are $\boldsymbol {\mu_B}$)

• How are you defining $\mu_b$? Because, I understand it to be $\frac{e\hbar}{2m}$. If that is the case, shouldn't $\mu=\sqrt 2 \mu_b$, rather than $\mu=\hbar \sqrt 2 \mu_b$ Commented Apr 5, 2015 at 19:34