# Finding the magnetic dipole moment In a Stern–Gerlach experiment

I'm trying to solve the following problem:

In a Stern–Gerlach experiment, the deflecting force on an atom is $F_z = -\mu_z\,{dB_z\over dz}$. In a particular experiment, the magnetic field region is $L = 50\,cm$ long, assume the magnetic field gradient is constant in that region, a beam of silver atoms enters the magnetic field with a speed of $v=375\, m/s$. What value of $dB_z/dz$ is required to give a separation of 1 mm between the two spin components as they exit the field?

My reasoning is as follows:

• Consider one silver atom with known mass m which is subjected to the magnetic force $F_z$, its motion is two-dimensional because the force deflects it from the straight path. Then applying Newton's second law in the x and z directions to obtain the trajectory equation.
• the maximum vertical deflection D is half of the separation between the two spin components, i.e $D = 0.5\,mm$.

The calculations are quite simple an elementary, it's completely analogous to the analysis of projectile motion in classical mechanics.
Newton's second law on z and x axes:
$$-\mu {dB_z\over dz} = ma_z$$ $$0 = ma_x$$ After integration and using of initial conditions:
$$z(t) = -{1\over 2}({\mu\over m}{dB_z\over dy})t^2$$ $$x(t) = vt$$
Solving this system of two equations by eliminating the time it gives us the trajectory equation of the atom:
$$z(x) = -({\mu\over 2v^2}{dB_z\over dz})x^2$$
Which is an equation of a parabola as expected (because the acceleration is constant).
Replacing z and x by D and L respectively, the equation becomes:
$$D = -({\mu\over 2v^2}{dB_z\over dz})L^2$$
Finally rearranging the equation and solve for the gradient ${dB_z\over dz}$.

Now I'm stuck with the z-component of the magnetic dipole moment $\mu_z$, I have no idea how to find its numerical value. I know there is an orbital magnetic moment $\mu_l$ and a spin magnetic moment $\mu_s$ of electrons, their z-component are given by: $$\mu_l= \mu_B g_l m_l$$ $$\mu_s= \mu_B g_s m_s$$

Where the g-factor is given by: $g_l = 1$ and $g_s = 2$.
$m_l$ and $m_s$ are the usual quantum numbers.

The problem is that I don't know to use these two formulae in this situation of many electrons.

So my question is:
How do I find $\mu_z$? Is the electronic configuration of the silver atom in the ground important here? What about the spin magnetic moment of the nucleus? Is it involved?

Any hint, idea or comment will be greatly appreciated.

• Is this of any use: hyperphysics.phy-astr.gsu.edu/hbase/spin.html there are lots of lecture note pdfs online.
– user154420
Jul 7 '17 at 14:24
• @Countto10 thanks! It's a very good article! it improved my understanding and even contains the solution of my problem!
– Samà
Jul 7 '17 at 15:02
• If you accept an answer, you can also upvote it, :) also, there are two books I would recommend for worked example after worked example, QM Demystified by McMahon and Squires Solved Problems in QM. No heavy duty blocks of texts, both stick to practical worked problems. Best of luck
– user154420
Jul 7 '17 at 15:14

The magnetic moment is given by source

$$\mu = \mu_B \frac{g}{2}$$

where $\mu_B=\frac{e\hbar}{2m_e}$ is the Bohr magneton, and $g$ is the electron spin g-factor (g=2.002319304386).

The reason that this is the only value you need to be concerned about is given on this page - last article:

The actual experiment was carried out with a beam of silver atoms from a hot oven because they could be readily detected using a photographic emulsion. The silver atoms allowed Stern and Gerlach to study the magnetic properties of a single electron because these atoms have a single outer electron which moves in the Coulomb potential caused by the 47 protons of the nucleus shielded by the 46 inner electrons. Since this electron has zero orbital angular momentum (orbital quantum number l=0), one would expect there to be no interaction with an external magnetic field.

So you don't have to worry about the orbital angular momentum, and only the spin of the electron itself matters.

• Thank you so much! So the orbital and spin magnetic dipoles of the 46 inner electrons are simply cancel out, and the only valence electron has a spin dipole only because it is an s electron (l=0). It makes sense.
– Samà
Jul 7 '17 at 15:07