Equations of motion of an electron emitted from a surface

In summary, the equations of motion of an electron emitted from a surface describe its trajectory and velocity as it interacts with external forces, such as electric and magnetic fields. The initial kinetic energy of the electron, determined by the work function of the material, influences its motion after emission. The equations incorporate concepts from classical mechanics and quantum mechanics, highlighting the effects of potential barriers and the influence of surrounding fields on the electron's path, ultimately providing insights into phenomena like photoelectric effect and electron dynamics in various environments.
  • #1
NB76
14
0
Homework Statement: Real world application of freshman physics
Relevant Equations: TBD

This is not a homework question, this is relevant to my work. It seems simple enough (introductory) but I keep running into problems.

An electron is emitted from an surface (material is irrelevant, could be an infinite pane, could be a disk, etc.). The plane is held at a potential V (V > 0, electron is attracted back to the surface). The initial kinetic energy is KE. The angle of emission is alpha. What are the equations of motion for the electron?
 
Physics news on Phys.org
  • #2
NB76 said:
could be an infinite pane, could be a disk, etc.
What is the field produced by an infinite charged plane?
 
  • #3
If there is a hint in there it is not obvious to me. What is the field form an infinite charged plane?

This is a real problem. The plane is actually 2 cm x 2 cm. I am trying to be flexible in my acceptance of how the problem is set up. If an infinite plane allows for a simple solution I want to see and understand that solution and it's limitation. If there is a better (simpler) geometry that would enable a simpler calculation I want to see and understand that solution and it's limitation.
 
  • #4
NB76 said:
If there is a hint in there it is not obvious to me. What is the field form an infinite charged plane?

This is a real problem. The plane is actually 2 cm x 2 cm. I am trying to be flexible in my acceptance of how the problem is set up. If an infinite plane allows for a simple solution I want to see and understand that solution and it's limitation. If there is a better (simpler) geometry that would enable a simpler calculation I want to see and understand that solution and it's limitation.
You mentioned infinite plane as a possibility, and that is the simplest case. The field is constant, so it will be a parabolic trajectory, just like a stone thrown in a local gravitational field.
Otherwise it will depend on how far the electron can get from the surface. If it can be a distance commensurate with the width of the surface then the field will diminish. When far away enough that the surface can be considered a point charge then it will be like a planet around the sun, following an ellipse or escaping completely.

So to answer the question in the thread's title, no, it isn't.
 
  • Like
Likes SammyS
  • #5
The problem is far from being simple. If you are looking for analitical equations of motion you will need to know the electric field in the space around your charged plate as a function of position. For an infinite plate the field is uniform in all space. For a finite plate is not. You may be able to approximate it with the field of an inifinite plate if the electron stays very close to the surface, compared with the plate size. If the electron moves over distances comparable with the size of the plate then you need to calculate the field numerically and then trace the motion of the electron also by numerical analysis.
On the other hand, if the electron is very far away from the plate, the field could be approximated with the field of a point charge.
But maybe you have two plates charged with opposte charges? When you say the the voltage of the plate is positive, this is relative to what? Is the electron moving between the plates of a capacitor?
 
  • #6
Thank you nasu and haruspex!

How far does the electron travel is the question I want to answer. If a first pass (infinite plane solution) tells me that the electron is only microns (?) off the surface, maybe the simplification (infinite plane) was justified. If that initial solution result in an answer of cm or meters than a more robust treatment is required. Can you get me started on the math for the infinite plane solution?

In reality, this is all done in a vacuum chamber, with the chamber wall at ground. Maybe a better approach would be treating the problem as a parallel plate capacitor where one plate it represents the sample and the other plate is the represents the chamber walls? Can you get me started on the math for the parallel plate capacitor solution?

I am new to the forum. Should I move this discussion to another, more advanced, section?
 
  • #7
NB76 said:
I am new to the forum. Should I move this discussion to another, more advanced, section?
No worries. I've gone ahead and moved the thread to the Classical Physics forum.
 
  • #8
You will need more information about the system. To estimate how far it goes you need to know the gradient of the potential and not just the potential of the plate (which is an arbitrary value anyway). Or the charge density on the plate.
 
  • #9
NB76 said:
this is relevant to my work.
Can you give a bit more context ?

NB76 said:
It seems simple enough (introductory) but I keep running into problems.
Can you describe those problems ?

NB76 said:
An electron is emitted from an surface (material is irrelevant, could be an infinite pane, could be a disk, etc.). The plane is held at a potential V (V > 0, electron is attracted back to the surface). The initial kinetic energy is KE. The angle of emission is alpha. What are the equations of motion for the electron?
If you express KE in eV, do you get something that is >> V , of the same order of magnitude as V , or << V ?

Equation of motion is ##\vec F = m\vec a## with ##\vec F = -e\vec E = e\vec\nabla V(x,y,z)## with ##V(x,y,z)## not the voltage of the plate but the electric potential field above the plate. See here for the integral

##m## is 9.109 ##\times 10^{-31}## kg
##e## is 1.602 ##\times 10^{-16}## C

Now, if I follow this thread, my estimate is that this doesn't help you very much. Right ? In that case we need more guidance from you to help us help you :smile:

##\ ##
 
  • #10
nasu said:
You will need more information about the system. To estimate how far it goes you need to know the gradient of the potential and not just the potential of the plate (which is an arbitrary value anyway). Or the charge density on the plate.
Thanks Nasu. I am not sure what you mean about the gradient on the plate. Let's say the plate is a conductor, held at fixed potential (relative to ground) with an external power supply. Does that mean the gradient is zero?
 
  • #11
BvU said:
Can you give a bit more context ?


Can you describe those problems ?


If you express KE in eV, do you get something that is >> V , of the same order of magnitude as V , or << V ?

Equation of motion is ##\vec F = m\vec a## with ##\vec F = -e\vec E = e\vec\nabla V(x,y,z)## with ##V(x,y,z)## not the voltage of the plate but the electric potential field above the plate. See here for the integral

##m## is 9.109 ##\times 10^{-31}## kg
##e## is 1.602 ##\times 10^{-16}## C

Now, if I follow this thread, my estimate is that this doesn't help you very much. Right ? In that case we need more guidance from you to help us help you :smile:

##\ ##
Hi BvU - Thanks for your input.

I am interested in electrons in the range of 2-20 eV. The potential on the plane could range form 0 - 100 V.
 
  • #12
NB76 said:
Thanks Nasu. I am not sure what you mean about the gradient on the plate. Let's say the plate is a conductor, held at fixed potential (relative to ground) with an external power supply. Does that mean the gradient is zero?
No, the gradient menas how the potential changes in the space near the plate. The value of the potential at the plate by itself is irelevant. You can make it zero or any value you want. Important is the change in potential. If the electron starts with some kinetic energy KE, it will stop when
##KE=q\DeltaV=q(V_{final-V_{plate})##
 
  • #13
How do I find the gradient?
 
  • #14
Here is a picture if that helps.
Electron_Problem.jpg
 
  • #15
NB76 said:
Here is a picture if that helps.
View attachment 342173
I answered this case in post #4. What is the force on the electron?
 
  • #16
@NB76 , let's take a step back. Before you find out the equations of motion, you need to know what problem you are solving. And the first question - which we still haven't answered - is whether the infinite plate approximation is valid or not.

Your electron starts with an initial kinetic energy T. It loses an energy qEh when it travels above the plane, where q is the electron charge, E is the electric field, and h is the height where it stops and starts its return. So h = T/qE. Now, if h is very small when compared to the size of the disk, you can treat it as an infinite plane. If h is very large when compared to the size of the disk, you can treat it as a point. In between, this gets very complicated.

So which is it?
 
  • #17
NB76 said:
How do I find the gradient?
This will depend on the details of your setup. You can't find it based on vague descriptions. You really need to specify the details of your setup. And eventually, what approximation will work, if any.
If it's an infinite plane, the field is uniform and the gradient of the potential is constant (and equal to the fieled, in magnitude).In this case, the motion is like a projectile motion. Same equations of motion,
 
  • #18
The negative charge of the emitted electron will attract positive free charge on the plane and cause some "bunching" of positive charge on the plane in the area beneath the electron. So, the electric field experienced by the electron will not be uniform. The electron will be attracted back toward the plane with a greater force than if the charge on the plane remained uniform. This effect is stronger when the electron is near the plane. Maybe this effect can be handled by introducing a positive image charge of the electron.
 
  • Like
Likes nasu
  • #19
Vanadium 50 said:
@NB76 , let's take a step back. Before you find out the equations of motion, you need to know what problem you are solving. And the first question - which we still haven't answered - is whether the infinite plate approximation is valid or not.

Your electron starts with an initial kinetic energy T. It loses an energy qEh when it travels above the plane, where q is the electron charge, E is the electric field, and h is the height where it stops and starts its return. So h = T/qE. Now, if h is very small when compared to the size of the disk, you can treat it as an infinite plane. If h is very large when compared to the size of the disk, you can treat it as a point. In between, this gets very complicated.

So which is it?
Sorry for the delayed response.

Assume r is the radius of the charged disk form which the electron is ejected (in reality it is a more complicated shape).

I don't know the answer. I want to solve the infinite plane problem first because it seems like the easiest problem. If the solution dictates that the electron does not travel far above (h << r) the plane then I know it was a reasonable assumption.

If the electron travels a great distance from from the plane (h >> r) I would model the plane as a point charge?

If h~=r then I have a much more difficult problem that has a non-trivial solution.

Does this approach seem reasonable?
 
  • #20
TSny said:
The negative charge of the emitted electron will attract positive free charge on the plane and cause some "bunching" of positive charge on the plane in the area beneath the electron. So, the electric field experienced by the electron will not be uniform. The electron will be attracted back toward the plane with a greater force than if the charge on the plane remained uniform. This effect is stronger when the electron is near the plane. Maybe this effect can be handled by introducing a positive image charge of the electron.
Could you give me a more detailed approach to starting this problem?
 
  • #21
NB76 said:
Sorry for the delayed response.

Assume r is the radius of the charged disk form which the electron is ejected (in reality it is a more complicated shape).

I don't know the answer. I want to solve the infinite plane problem first because it seems like the easiest problem. If the solution dictates that the electron does not travel far above (h << r) the plane then I know it was a reasonable assumption.

If the electron travels a great distance from from the plane (h >> r) I would model the plane as a point charge?

If h~=r then I have a much more difficult problem that has a non-trivial solution.

Does this approach seem reasonable?
Looks good to me.

Remember that a uniformly charged sphere or spherical shell will produce the same field as a point charge, so that approximation might be usable for ##h\approx r## too.
Calculate the value of h at which the sheet and the point would produce the same field, then use the sheet approximation up to that distance.

I think @TSny's point about a changed distribution on the sheet is only relevant if the initial charge density and velocity are very low. (And certainly only if it is a conductor, which you have not mentioned.) E.g. if the sheet is an uncharged conductor then it is the same as an electron and a positron moving apart in mirror image trajectories.
 
  • #22
NB76 said:
Here is a picture if that helps.
View attachment 342173
Just to reiterate where I am in this problem. I know the field from a 2-D infinite plane is E = 2 k pi sigma. I don't know how to calculate sigma for an infinite plane held at a fixed voltage V. Can anyone assist with this?

With the field quantified, I expect the y-position to vary proportional to E t^2 (third equation, second term). Can I get some assistance with this second term?
 
  • #23
haruspex said:
Looks good to me.

Remember that a uniformly charged sphere or spherical shell will produce the same field as a point charge, so that approximation might be usable for ##h\approx r## too.
Calculate the value of h at which the sheet and the point would produce the same field, then use the sheet approximation up to that distance.

I think @TSny's point about a changed distribution on the sheet is only relevant if the initial charge density and velocity are very low. (And certainly only if it is a conductor, which you have not mentioned.) E.g. if the sheet is an uncharged conductor then it is the same as an electron and a positron moving apart in mirror image trajectories.
Thanks NB76 - could you help me put this into a formulaic form?

Just to reiterate where I am in this problem. I know the field from a 2-D infinite plane is E = 2 k pi sigma. I don't know how to calculate sigma for an infinite plane held at a fixed voltage V. Can anyone assist with this?

With the field quantified, I expect the y-position to vary proportional to E t^2 (third equation, second term in the diagram). Can I get some assistance with this second term?
 
  • #24
NB76 said:
Thanks NB76 - could you help me put this into a formulaic form?

Just to reiterate where I am in this problem. I know the field from a 2-D infinite plane is E = 2 k pi sigma. I don't know how to calculate sigma for an infinite plane held at a fixed voltage V. Can anyone assist with this?

With the field quantified, I expect the y-position to vary proportional to E t^2 (third equation, second term in the diagram). Can I get some assistance with this second term?
Using the formula for the potential at distance x from the centre of a finite uniformly charged disc (https://phys.uri.edu/gerhard/PHY204/tsl82.pdf), obtain an approximation for small x. You should then be able to relate the potential at distance x to the potential at the centre of the disc.
 
  • #25
haruspex said:
Using the formula for the potential at distance x from the centre of a finite uniformly charged disc (https://phys.uri.edu/gerhard/PHY204/tsl82.pdf), obtain an approximation for small x. You should then be able to relate the potential at distance x to the potential at the centre of the disc.
I think I am asking a different question. I know the potential on the plane. I hook a power supply to the plane and set a voltage. I don't know the charge on the plane (sigma). How do I get from V (power supply) to sigma?
 
  • #26
You can't connect the power supply just to the plane. You need to connect to something else the other terminal of the power supply. The charge on the plane will depend on the capacitance of the system plane + the something else.or the plane and the "ground".
 
  • #27
NB76 said:
I think I am asking a different question. I know the potential on the plane. I hook a power supply to the plane and set a voltage. I don't know the charge on the plane (sigma). How do I get from V (power supply) to sigma?
The power supply will be referenced to ground (see drawing).
 
  • #28
NB76 said:
I think I am asking a different question. I know the potential on the plane. I hook a power supply to the plane and set a voltage. I don't know the charge on the plane (sigma). How do I get from V (power supply) to sigma?
What I wrote gives you a way to relate the potential of the disc to its charge density.
 
  • #29
The drawing is just a schetch of the principle. The actual charge desnity will depend on the actual sizes and shapes of the components. You cannot get it from general principles. It's the same idea as knowing the actual map of the potential in the system. If you know how thw potential changes from point to point you don't need the charge density.
 
  • #30
nasu said:
If you know how thw potential changes from point to point
… which we don't without doing something equivalent to what I suggested.
 
  • #31
haruspex said:
What I wrote gives you a way to relate the potential of the disc to its charge density.
I am sorry, I just don't see it (get it).
 
  • #32
NB76 said:
I am sorry, I just don't see it (get it).
##V(x)=2\pi \sigma k(\sqrt{R^2+x^2}-x)\approx 2\pi \sigma kR(1-\frac xR)=2\frac{Qk}R(1-\frac xR)=V_0(1-\frac xR)##.
Equating that to the large x approximation, ##\frac{Qk}x##, gives a quadratic for estimating where you should switch from one approximation to the other.
 
  • #33
The equations of motion for an electron emitted with an initial kinetic energy \(KE\) at an angle \(\alpha\) to the surface can be derived from the laws of conservation of energy and momentum. Since the electron is emitted with an initial speed \(v_0\), its initial kinetic energy is \(KE = \frac{1}{2} m v_0^2\), where \(m\) is the mass of the electron. Under the influence of the electric field created by the potential \(V\), the electron will change its speed and direction of movement. Taking into account the law of conservation of energy (\(E = KE + eV\), where \(e\) is the charge of the electron) and the law of conservation of momentum (\(mv_0 = mv + meV\)), we can obtain equations describing the trajectory of the electron.
 
  • #34
What is the meaning of the meV term in the "conservation of momentum"? Is this a quantity with dimensions of momentum?
 

FAQ: Equations of motion of an electron emitted from a surface

What are the basic equations of motion for an electron emitted from a surface?

The basic equations of motion for an electron emitted from a surface can be derived from classical mechanics, specifically Newton's second law, and can be expressed as:

F = ma, where F is the net force acting on the electron, m is the mass of the electron, and a is its acceleration. The motion can also be described using kinematic equations, such as:

x = x₀ + v₀t + (1/2)at², where x is the final position, x₀ is the initial position, v₀ is the initial velocity, and t is time.

What factors influence the motion of an emitted electron?

The motion of an emitted electron is influenced by several factors, including the initial kinetic energy of the electron, the electric and magnetic fields present in the environment, and interactions with other particles or surfaces. The work function of the material from which the electron is emitted also plays a crucial role in determining the initial energy of the electron.

How does the electric field affect the motion of an emitted electron?

An electric field exerts a force on the emitted electron, causing it to accelerate. The force experienced by the electron can be calculated using F = qE, where q is the charge of the electron and E is the electric field strength. This acceleration affects the trajectory of the electron, causing it to move in a direction influenced by the orientation of the electric field.

Can the motion of an emitted electron be described using quantum mechanics?

Yes, the motion of an emitted electron can also be described using quantum mechanics. When considering phenomena such as tunneling or the discrete energy levels of electrons in atoms, quantum mechanics provides a more accurate framework. The wave function of the electron can be used to describe its probability distribution and behavior, which differs from classical trajectories.

What is the significance of the emission angle of an electron?

The emission angle of an electron is significant as it can influence the electron's subsequent trajectory and interactions with external fields. Depending on the angle at which an electron is emitted, it may encounter different forces and potentials, affecting its energy loss and scattering processes. This angle is often determined by the surface properties and the energy distribution of the emitted electrons.

Back
Top