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

Click For Summary
SUMMARY

The discussion focuses on determining the field gradient of a 50 cm long Stern-Gerlach magnet required to achieve a 1 mm separation between spin-up and spin-down silver atoms emitted from an oven at 1500 K. The average kinetic energy of the emitted atoms is calculated as 2kBT, where kB is the Boltzmann constant. The confusion arises from interpreting the formula involving the terms a_z (l_1/v) and a_z t_1, which represent the distance traveled within and outside the magnet, respectively. The solution clarifies that the second term accounts for the constant velocity outside the magnet, leading to a comprehensive understanding of the kinematic equations involved.

PREREQUISITES
  • Understanding of Stern-Gerlach experiments and their principles
  • Familiarity with kinematic equations in physics
  • Knowledge of thermal physics, specifically kinetic energy calculations
  • Basic understanding of magnetic dipole moments and the Bohr magneton
NEXT STEPS
  • Study the principles of the Stern-Gerlach experiment in detail
  • Learn about kinematic equations and their applications in physics
  • Research the significance of the Boltzmann constant in thermal physics
  • Explore the concept of magnetic dipole moments and their calculations
USEFUL FOR

Physics students, educators, and researchers interested in quantum mechanics, particularly those studying atomic behavior in magnetic fields and the Stern-Gerlach experiment.

grandpa2390
Messages
473
Reaction score
14

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
 

Attachments

  • Screen Shot 2017-09-25 at 10.56.50 AM.png
    Screen Shot 2017-09-25 at 10.56.50 AM.png
    27.7 KB · Views: 921
Physics news on Phys.org
Can you interpret the quantity ##a_z \left(\frac{l_1}{v} \right) = a_z t_1##?
 
  • Like
Likes   Reactions: grandpa2390
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.
 
  • Like
Likes   Reactions: grandpa2390

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 32 ·
2
Replies
32
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K