Townsend - Problem 1.1

Chapter 1 - Stern Gerlach Experiments

Problem 1.1
Determine the field gradient of a 50-cm-long Stern Gerlach magnet that would produce a 1-mm separation at the detector between spin-up and spin-down silver atoms that are emitted from an oven at $T = 1500 K$. Assume the detector is located 50 cm from the magnet. Note that the emitted atoms have average kinetic energy $2k_bT$.

Solution
We are interested in the field gradient necessary to produce a given pattern on the detector, hence we need to find $\frac{\partial B}{\partial z}$.

From equation 1.4 we have that the force due to a Stern-Gerlach magnet is given by
$$F_z = \mu_z \frac{\partial B}{\partial z}$$,
where $\mu_z$ is the magnetic dipole moment of the particle given by
$$\mu_z = \frac{gq}{2mc}S_z$$ with $S_z$ being the intrinsic spin angular momentum of the particle in the z-direction.

For silver atoms, the magnetic dipole moment is due to the intrinsic spin of a single electron. Hence $g = 2$ and $q = e$, the charge of an electron.

Notice that, using symmetry arguments, the displacement in the z-direction of the upper particle is the same as the displacement of the lower particle. Let this displacement be $z$. Then $d = 1 mm = 2z$.

Therefore, we can focus on the displacement in the z-direction of the upper particle.
While the particle is traveling through the magnet it experiences a force and thus accelerates in the z-direction. We have
$$a_z = \frac{F_z}/{M} = \frac{\mu_z}/{M} \frac{\partial B}{\partial z}$$ where $M$ is the mass of the silver atom.

When the magnet is not traveling through the magnet it does not accelerate in the z-direction.

Initially there is no velocity in the z-direction and we have that the displacement in the z-direction of the upper particle while it is traveling through the magnet is given by
$$z_1 = \frac{1}{2}a_zt^2 = \frac{1}{2}a_z \left(\frac{l}{v_x}\right)^2$$, where $v_x$ is the initial velocity of the particle in the x-direction.

Outside the magnet the displacement in the z-direction of the upper particle is given by:
$$z_2 = v_zt = v_z\frac{l}{v_x} = a_z\left(\frac{l}{v_x}\right)^2$$
because v_z is the velocity of the particle when it just left the magnet.

The total displacement is thus
$$z = z_1 + z_2 =  \frac{3}{2}a_z\left(\frac{l}{v_x}\right)^2$$

We can solve for $v_x$ by making use of the fact that the particles are ejected with average kinetic energy $2k_bT$.
$$\frac{1}{2}M{v_x}^2 = 2k_bT$$
Thus  ${v_x}^2 = \frac{4k_bT}{M}$ and we can solve

$$d = 2z = 3a_zl^2\frac{M}{4k_bT} =  \frac{3}{4}\frac{\mu_z}{M} \frac{\partial B}{\partial z}\frac{4k_bTl^2}{M}$$

Finally, we rearrange for $\frac{\partial B}{\partial z}$ and plug in the values:

• d = 0.1 cm
• l = 50 cm
• T = 1500  

to obtain
$$\frac{\partial B}{\partial z} = 1200 G/cm$$