Stern-Gerlach Separation Distance on Target Film

[Quelsita]

I have a question in which we are instructed to find the separation distance between the two impact zones of Sodium atoms on a photographic plate:


A magnet of length l= 0.15m and field gradient $$\nabla$$B= 9.0T/m over its full length is used in a Stern-Gerlach experiment. If a beam of Sodium atoms of velocity v=600m/s passes through this magnet and then strikes a photographic plate at a distance of d=0.80 m beyond the magnet, what will the separation between the two impact zones be on the film?
Sodium has total-angular momentum of J=1/2, contributed by the spin of an electron of the 3s subshell.

I understand that as the electrons enter the magnet, half of the electrons will experience spin-up or spin-down depending if s is positive or negative and thus, there are only two permitted magnetic moments which gives two impact zones on the photo film.
However, I am confused as to how to find the distance between the two impact zones based on the information given.

Any assistance is welcome!
Thanks!

 

[jbsteele]

The separation distance between the two impact zones is given by the formula: d = (\nabla B * l^2)/(2mv), where m is the mass of the Sodium atom, v is its velocity, l is the length of the magnet, and \nablaB is the field gradient of the magnet. Using the given values for each of these parameters, we have: d = (9.0 T/m * 0.15m^2)/(2 * (22.98977 * 10-3 kg)*600m/s) = 0.0047m. Therefore, the separation distance between the two impact zones on the photographic plate is 0.0047m.

 

[nlightner]

Based on the information provided, we can calculate the magnetic moment of Sodium atoms using the total-angular momentum and the Bohr magneton. The magnetic moment is given by μ = JgμB, where J is the total-angular momentum, g is the Lande g-factor, and μB is the Bohr magneton.

For Sodium, g = 2 and μB = 9.274 x 10^-24 J/T. Therefore, the magnetic moment of Sodium atoms is μ = 1/2 x 2 x 9.274 x 10^-24 J/T = 9.274 x 10^-24 J/T.

Next, we can use the equation for the deflection of a particle in a magnetic field to find the separation distance between the two impact zones on the photographic plate. This equation is given by d = μv/2B, where d is the separation distance, μ is the magnetic moment, v is the velocity of the particle, and B is the magnetic field gradient.

Substituting the given values, we get d = (9.274 x 10^-24 J/T)(600 m/s)/(2 x 9.0 T/m) = 3.09 x 10^-5 m.

Therefore, the separation distance between the two impact zones on the photographic plate will be approximately 3.09 x 10^-5 m. I hope this helps clarify your confusion.