Why canal rays need a perforated cathode?

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Discussion Overview

The discussion centers on the necessity of a perforated cathode in experiments involving canal rays and cathode rays, exploring the mechanisms of fluorescence production and the behavior of positive ions versus electrons in a vacuum tube environment.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions why canal rays require a perforated cathode to produce fluorescence on the opposite side, unlike cathode rays which can produce fluorescence near the anode without a perforation.
  • Another participant explains that canal rays are positive ions and cathode rays are negative ions, suggesting that the thickness of the cathode prevents both types of rays from passing through without holes.
  • It is noted that the fluorescence observed is primarily due to scattered electrons, which are lighter and scatter more easily than the heavier positive ions.
  • A participant expresses confusion about the scattering of cathode rays, questioning whether they would be attracted to the anode instead of scattering upon impact.
  • Another reply indicates that the energy of the particles affects scattering, with higher energy leading to increased scattering.

Areas of Agreement / Disagreement

Participants express differing views on the behavior of canal rays and cathode rays, particularly regarding the necessity of a perforated cathode and the mechanisms of fluorescence production. The discussion remains unresolved with multiple competing explanations.

Contextual Notes

Some assumptions about the thickness of the cathode and the nature of particle interactions are not fully explored, and the discussion does not clarify the specific conditions under which scattering occurs.

ananthu
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In the cathode ray production experiment, at about .01 mm of Hg, cathode rays travel from cathode and move towards anode and produce fluorescence on the walls near the anode.At the same time canal rays also move towards cathode. But only when you use a perforated cathode, the canal rays penetrate through it and produce fluorescence at the back of the cathode. I have doubts on the following aspects of the above experiment.
We are not using any perforated anode in the case of cathode ray production. Still these rays are able to produce fluorescence on the walls near the anode, thus indicating their presence. But why, the canal rays or the anode rays, so to speak, do need a perforated cathode to reach the other side and produce glow on the fluorescence screen kept at the back of the perforated cathode? Can not the anode rays make the walls near the cathode, glow even without a perforation in the cathode, as the cathode rays did?

As I am not able find the reason for it, it will be of great help to me if some experts on the subject throw some light on this aspect.
 
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Canal rays are positive ions, while cathode rays are electrons or other negative ions.

None of them make it through if there are no holes ... the anode and the cathode are usually much too thick for particles to pass through; even a micron is too thick!

The fluorescence you see on the walls is from scattered electrons; they are much lighter than the positive ions, and hence scatter much more easily.
 
Thanks for the reply.Sorry for my delayed response.

UltrafastPED said:
Canal rays are positive ions, while cathode rays are electrons or other negative ions.

The fluorescence you see on the walls is from scattered electrons; they are much lighter than the positive ions, and hence scatter much more easily.

Regarding the above statement, I could understand the light rays being scattered by an object. But how does it happen in the case of cathode rays? The latter being negative, when they hit the positive anode, will they not be attracted towards or absorbed by the anode rather than getting scattered?
 
Depends on the energy - if the particle energy is low enough there will be minimal scattering - but as the particle energy increases more and more scattering will be observed.
 

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