Determing field gradient of Stern-Gerlach magnet

[jay ess]

Homework Statement

Determine the field gradient of a 50-cm-long Stern-Gerlach magnet (d₁) 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. Assume the detector is located 50 cm from the magnet (d₂). Note: While the atoms in the oven have average kinetic energy 3k_BT2, the more energetic atoms strike the hole in the oven more frequently. Thus the emitted atoms have average kinetic energy 2k_BT2, where the k_B is the Boltzmann constant. The magnetic dipole moment of the silver atom is due to the intrinsic spin of the single electron. The Bohr magneton, e[STRIKE]h[/STRIKE]/2m_ec≈5.788×10^-9eV/G

Homework Equations

F_z=μ⋅∂B/∂z≈μ∂B_z/∂z

The Attempt at a Solution

The setup: the magnetic field gradient is oriented in the z-direction while the initial velocity of the atoms is in the x-direction.
The separation between the silver atoms is 1-mm, therefore, the distance traveled in the z-direction by the silver atoms is 0.5 mm=5x10^-2cm which I call d_z.
∂B_z/∂z=∇B
F_z=ma_z=μ∂B_z/∂z=μ∇B
Average kinetic energy of the particles: 1/2mv_x²=2k_BT→v_x=√(4k_BT/m)
v_x=d₁/t₁→t₁=d₁/v_x=d₁√(m/4k_BT)
I know that while the atoms are in the field gradient they will be experiencing a force that causes them to move in the z-direction (I'm just sticking with the positive z-direction for simplicity) and once they're out of the field gradient they will be traveling at a constant velocity in both the x- and z-directions. I also know that I need to somehow relate all of this to the kinematic equations but I'm kind of at a loss right now.
v_z=a_zt₁ (since there's no initial velocity in the z-direction and acceleration is constant).

 

[TSny]

Hi, Jay ess.

You've got a good start. Can you find an expression for the z- deflection of an atom while traveling through the magnet in terms of t₁? As the atom leaves the magnet it will have a z-component of velocity as you indicated as well as the x-component. Can you use these two components of velocity to determine the direction that the atom is traveling as it leaves the magnet? Then see if you can find the additional z-deflection as it travels to the detector.

 

[jay ess]

That's the bit that's giving me the most trouble, relating the kinematics equations in all of this mess. I know that since the atom started out at the origin and without any initial velocity in the z-direction that I can find the position of the particle as it is leaving the gradient (let's call it d_z1) by d_z1=1/2a_z*t₁²=1/2μ∇B/m(m/4k_BT)=μ∇B/(8k_BT). But, after that I'm still kind of lost because everything else I try just seems to fall apart.

 

[TSny]

d_z1=1/2a_z*t₁²=1/2μ∇B/m(m/4k_BT)=μ∇B/(8k_BT).

That look's good except for leaving out a factor of d₁².

But, after that I'm still kind of lost because everything else I try just seems to fall apart.

OK. You just have to calculate the additional displacement in the z direction after the atom leaves the magnet. This will be determined by the z-component of velocity and the time to get from the magnet to the detector. Since no force acts on the atom as it travels to the detector, the z-component of velocity will be constant and will equal whatever the z-component of velocity was at the instant the atom emerged from the magnet.