Ring or the shell in Bremsstrahlung?

In summary, the video explains how Thermal Bremsstrahlung works by dividing the set of electrons up into two sets of velocities, summing them over all different velocity directions, and using the result to determine the number density of electrons. The video also explains why a ring is a better representation of the problem than a shell. However, the video does not explain the why all the electrons inside the ring are assumed to have the same parameter b.
  • #1
tze liu
57
0
Recently i watch a video about Thermal Bremsstrahlung.
and i don't understand the explanation that they use a ring instead of disk in 6:02 to 7:06.

(1)how did he divided the set of electrons up into in
two sets of velocities that are all in
the same direction and sum
over all of the different velocity directions?

(2)why is this related to the reason he uses a ring instead of a shell?

the another question is even there are some electrons(ne) stayed inside the area between b and b+db
this doesn't mean those electrons are in the distance of closest appraoch,and this only means some of them pass through this area(may be they can appraoch the nucleus which has radius r < b),and i get stuck why this logic works here.I am not good at understand those basic concept about thermal free-free emission.

thank you very much!

-----------------
6:02
they might wonder why we've chosen to
06:03
examine a ring here instead of a shell
06:05
and the answer is a little hard to
06:07
visualize but we'll give it a shot we're
06:09
starting with a number density of
06:10
electrons we're asking how many of them
06:13
passed by this atom at a distance B if
06:16
we selected only the electrons out of
06:19
this cloud that were moving say from the
06:21
left side of the page to the right side
06:22
of the page then for that fixed
06:24
direction we would only see a mission
06:26
from these electrons in one ring as it
06:30
passes by the nucleus similarly if we
06:33
start if we selected only the electrons
06:35
that were starting at the bottom of the
06:36
page and moving up towards the top those
06:38
electrons also would only emit towards
06:41
as they pass through this ring so if you
06:44
divide your set of electrons up into in
06:47
two sets of velocities that are all in
06:49
the same direction so all the electrons
06:50
that are going in a similar direction
06:52
each one of those is only emitting over
06:55
this region that that's a ring so when
06:58
we sum over all of the different
06:59
velocity directions that these electrons
07:01
are traveling in multiplied by the
07:04
factor of the differential ring when we
07:07
add those all together it's equal to the
07:09
number density of the electrons times
07:11
the same ring size for all of those
07:13
different velocity directions so that's
07:14
why we end up with only a factor of a
07:16
ring and not a shell

 
Last edited:
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  • #2
I believe the answer may be that a ring is a 2-D representation of the problem as opposed to a shell, which is a 3-D representation. In looking at the ring, one is looking at the radial dependence (r) of the problem without consideration of the azimuth. Electron shells are idealizations of a complicated concept. When one talks of an orbital radius, one is referring to the most probable radial distance that an electron would be observed with respect to the center of the atom/nucleus.

I haven't reviewed the whole video, so hopefully others who have will respond.
 
  • Like
Likes tze liu
  • #3
There is one problem.
Even they use a ring instead of a shell,this doesn't means all electrons inside the ring has a parameter b respect to the center.
Why all the electrons inside the ring are assumed to have the same parameter b here?
 
  • #4
That's the definition of the ring.
You have to integrate over all rings later if you want to consider the total interaction.
 
  • #5
mfb said:
That's the definition of the ring.
You have to integrate over all rings later if you want to consider the total interaction.
i don't understand the detail of the explanation in the video
unfortunately

it is why i get stuck :(
 

Related to Ring or the shell in Bremsstrahlung?

1. What is the "ring" or "shell" in Bremsstrahlung?

The "ring" or "shell" in Bremsstrahlung refers to the characteristic emission spectrum produced when electrons are decelerated by a heavy nucleus. This produces a continuum of radiation that is typically described as a "ring" or "shell" shape in diagrams.

2. How is the "ring" or "shell" in Bremsstrahlung created?

The "ring" or "shell" in Bremsstrahlung is created when electrons are accelerated and then decelerated by a heavy nucleus, such as in an x-ray tube. As the electrons are decelerated, they emit a continuum of radiation, resulting in the characteristic "ring" or "shell" shape in the emission spectrum.

3. What is the significance of the "ring" or "shell" in Bremsstrahlung?

The "ring" or "shell" in Bremsstrahlung is significant because it allows scientists to identify and measure the energy of decelerated electrons. This emission spectrum is used in x-ray imaging and other applications to produce images of structures within the body.

4. How does the energy of the "ring" or "shell" in Bremsstrahlung relate to the energy of the decelerated electrons?

The energy of the "ring" or "shell" in Bremsstrahlung is directly related to the energy of the decelerated electrons. The higher the energy of the electrons, the higher the energy of the emitted radiation, resulting in a broader and brighter "ring" or "shell" in the emission spectrum.

5. Can the "ring" or "shell" in Bremsstrahlung be used to identify different elements?

Yes, the "ring" or "shell" in Bremsstrahlung can be used to identify different elements. Each element has a unique emission spectrum, and the characteristic "ring" or "shell" shape can help scientists distinguish between different elements and their energy levels.

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