Determine Field Gradient of a Stern Gerlach Magnet....

AI Thread Summary
The discussion centers on determining the field gradient of a Stern-Gerlach magnet to achieve a specific separation of silver atoms based on their spin states. The problem involves understanding the kinematic equations and the average kinetic energy of the emitted atoms. A key point of confusion is the interpretation of the formula involving acceleration and time, specifically the term a_z(t1), which represents the distance traveled outside the magnet after the atoms exit. The first term in the equation accounts for the distance traveled within the magnet, while the second term describes the motion outside the magnet at constant velocity. Clarifying these concepts helps in solving the problem effectively.
grandpa2390
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Homework Statement



Here is the Problem. I have the Solution but am having trouble understanding parts of it.

1.1. Determine the field gradient of a 50-em-long Stem-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 (see Fig. 1.1) is located 50 em from the magnet. Note: While the atoms in the oven have average kinetic energy 3kBT j2, the more energetic atoms strike the hole in the oven more frequently. Thus the emitted atoms have average kinetic energy 2kBT, where kB is the Boltzmann constant. The magnetic dipole moment of the silver atom is due to the intrinsic spin of the single electron. Appendix F gives the numerical value of the Bohr magneton, in a convenient form.

Homework Equations


Kinematic Equations

The Attempt at a Solution


The one part I don't understand is this:
?temp_hash=f22fd7ba1bdc435f42600ce1fe9722d1.png


I get that the units work out. But I don't understand why the formula is in this way. particularly the second term that isn't 1/2at^2 but a*t1*t2
 

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Can you interpret the quantity ##a_z \left(\frac{l_1}{v} \right) = a_z t_1##?
 
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TSny said:
Can you interpret the quantity ##a_z \left(\frac{l_1}{v} \right) = a_z t_1##?

As always. I stood stumped about the question for days. when I finally ask for help, it clicks.

the first term is the distance traveled up in the magnets.
the second term is the distance traveled up outside the magnets. there is no acceleration it is just vt where v is the velocity after leaving the magnet and t is the time spent outside of the magnet. : )
 
Yes. Good.
 
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