Cyclotron Frequency of a Hydrogen Atom in a Field B

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    Cyclotron Frequency
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SUMMARY

The discussion focuses on the cyclotron frequency of a hydrogen atom in a magnetic field B, specifically analyzing the torque on the electron and its relationship to angular frequency. The derived equations are \(\omega_p = \frac{gqB}{2m}\) for the torque and \(\omega = \frac{qB}{m}\) for a single electron's motion. The speaker notes that while these two cases yield similar results, differing by a factor of 2, they seek to understand the underlying connection or coincidence between them. The reference to Feynman's lectures indicates a foundational approach to the topic.

PREREQUISITES
  • Understanding of classical electromagnetism principles
  • Familiarity with angular momentum and torque concepts
  • Knowledge of the hydrogen atom's electron dynamics
  • Basic grasp of Feynman's lectures on quantum mechanics
NEXT STEPS
  • Study the derivation of cyclotron frequency in different contexts
  • Explore the implications of torque on electron motion in magnetic fields
  • Investigate the role of spin in the behavior of electrons in magnetic fields
  • Review Feynman's Lectures on Physics, particularly Volume VII, Section 34-3
USEFUL FOR

Physics students, researchers in quantum mechanics, and educators looking to deepen their understanding of electron dynamics in magnetic fields.

jackychenp
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Hi All,

For a hydrogen atom in the field B, torque on the electron is [itex]\tau =\mu Bsin\theta =\omega_pJsin\theta[/itex] => [itex]\omega_p=gqB/(2m)[/itex] If we only consider orbital motion (ignore spin), then g=1. So [itex]\omega_p=qB/(2m)[/itex]. (It follows the steps from Feynman's lectures VII section 34-3)
For a single electron in the field, [itex]qvB=v^2m/r[/itex] => [itex]\omega=qB/m[/itex]. I know these two cases are different, but why the results look the same (only a factor of 2 different). Is there any connection between them, or just coincidence?
 
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Getting the dimensions to come out right requires the ratio qB/m.
 

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