Bohr's calculation of stopping power

In summary: E_perpendicular is in SI units, but qenc is not. I'm not sure where the 4π comes from. :smile: No, I mean I know Coulomb's law, but the four pi seems to come out of nowhere when you consider that Gauss' law:ε0∫E . dA = qenc
  • #1
EucharisCriss
2
0
so, hey...I'm pretty new to these forums and just needed help understanding a textbook that I'm reading.

I'll give you all the background, but I'm not sure if it is necessary.

They begin by considering a heavy particle of mass M and velocity v passing through a material with an atomic electron at distance b from the particle trajectory. the electron is assumed to be free and initially at rest. Furthermore, it is assumed that the electron only moves slightly during the interaction with the heavy particle. After the collision, it is assumed that the incident heavy particle continues essentially as before (because M>>me)

the impulse is calculated:
I=integral(Fdt)=e * integral(E_perpendicular)[tex]\frac{dt}{dx}[/tex]dx = e integral ( E_perpendicular [tex]\frac{dx}{v}[/tex]

They claim that only E_perpendicular enters because of symmetry. I am a little confused about this actually. I kind of reasoned it out by thinking of the work function (which would give the change in energy for the particle, right) W=integral(F . dx )=integral (eE . dx) = integral( e*E_perpendicular*dx)...(their next to last step seems kind of like this...)...but I think I'm just trying to make up something to understand it...so if anyone could tell me what symmetry they mean, I'd appreciate it.

Also, they go on to calculate the integral( E_perpendicular dx) using Gauss' Law over an infinitely long cylinder centered on the particle trajectory. They claim that:

integral ( E_perpendicular *2*pi*b*dx)=4*pi*z*e

I just don't know where the 4*pi is coming from on the right side (and where epsilon is..) and would really appreciate it if someone could explain this to me.

I hope this is an okay place to post this. It isn't homework, I'm just reading it...

I tried to draw the diagram from the book and it is attached (i hope).

I just feel like I must be missing something major
 

Attachments

  • diagram2.bmp
    245.5 KB · Views: 649
Physics news on Phys.org
  • #2
Welcome to PF!

Hi EucharisCriss! Welcome to PF! :smile:

(have an integral: ∫ and a pi: π and an epsilon: ε :wink:)
EucharisCriss said:
They claim that only E_perpendicular enters because of symmetry. I am a little confused about this actually.

I think the symmetry is because the effect before M reaches the nearest point will be the reflection of the effect after, so the total effect will be perpendicular. :wink:
Also, they go on to calculate the integral( E_perpendicular dx) using Gauss' Law over an infinitely long cylinder centered on the particle trajectory. They claim that:

integral ( E_perpendicular *2*pi*b*dx)=4*pi*z*e

I just don't know where the 4*pi is coming from on the right side (and where epsilon is..) and would really appreciate it if someone could explain this to me.

4πε0 is the factor you always need to include when you use SI units.

(And ε0 is the permittivity of the vacuum)

See the PF Library on Coulomb's law for details. :smile:
 
  • #3
No, I mean I know Coulomb's law, but the four pi seems to come out of nowhere when you consider that Gauss' law:

ε0∫E . dA = qenc

I mean, Coulomb's law can be derived from this by considering a point charge with a spherical gaussian surface centered on the particle.
ε0∫E . dA = qenc
ε0∫E dA = qenc (because E and dA are at 0 degree angle when charge is positive)
ε0EA = qenc
In this case A=4πr^2
so this is where the 4πε0 in Coulomb's law comes from...

However, in their version of gauss' law, they have no ε0 and an extra 4π on the side side of the enclosed charge (the enclosed charge being ze).
They have changed dA to 2πbdx (2πb being the circumference of the circle and dx being the length of the differential cylindrical shell).
Basically, the left side of this equation makes sense while the existence of the 4π (and the lack of an ε0) on the right sense seems to come from nowhere for me.

What I was expecting to see:
rightside=ze/ε0
What they have:
rightside=4πze

ε0 is slightly related to 4π because c=1/sqrt(ε0μ0)
μ0=4π * 10 ^-7


I would agree with you about the symmetry argument, but it was my thought that the charged particle ze is on a collision path with the electron. I should get the book out, but I'm kind of lazy right now.
 
  • #4
EucharisCriss said:
… However, in their version of gauss' law, they have no ε0 and an extra 4π on the side side of the enclosed charge (the enclosed charge being ze). …

Hi EucharisCriss! :smile:

My guess is that the book isn't using SI units.
 

1. What is Bohr's calculation of stopping power?

Bohr's calculation of stopping power is a mathematical formula that predicts the amount of energy lost by a charged particle as it passes through a material. It takes into account the properties of the particle and the material it is passing through.

2. How did Bohr develop this calculation?

Bohr developed his calculation by building on the work of other scientists, including Ernest Rutherford and Niels Bohr. He used principles of classical mechanics and quantum mechanics to develop a more accurate and comprehensive formula.

3. What factors does Bohr's calculation take into account?

Bohr's calculation takes into account the energy and charge of the particle, as well as the atomic number and density of the material it is passing through. It also considers the velocity of the particle and the distance it travels through the material.

4. What is the significance of Bohr's calculation of stopping power?

Bohr's calculation of stopping power is significant because it allows scientists to accurately predict the behavior of charged particles as they pass through different materials. This is important in fields such as nuclear physics, medical physics, and radiation therapy.

5. Are there any limitations to Bohr's calculation?

While Bohr's calculation of stopping power is a highly accurate and useful tool, it does have some limitations. It is most accurate for low-energy particles and can become less accurate for higher energies. It also does not take into account certain factors such as the magnetic field or spin of the particle.

Similar threads

Replies
68
Views
3K
Replies
4
Views
906
Replies
2
Views
136
Replies
11
Views
1K
Replies
4
Views
143
  • Beyond the Standard Models
Replies
0
Views
918
  • Calculus and Beyond Homework Help
Replies
9
Views
686
  • Mechanics
Replies
1
Views
964
Back
Top