Finding normalization constant for proton charge distributio

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The discussion revolves around finding the normalization constant for the proton's charge distribution. The user initially misapplies concepts from quantum mechanics, confusing charge density with wave functions. Key insights reveal that the charge density must be integrated over volume to equal the total charge of the proton, which is e (1.6×10^-19 C). The correct volume element in spherical coordinates is identified, and the user realizes that the integral should equal the total charge rather than 1. Ultimately, the normalization constant is derived as N = 1/(4*pi*R^2) to ensure the charge distribution integrates to the total charge of the proton.
ElectricEel1
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Homework Statement

For a model of a proton's charge distribution,
hw1eqn2.gif
:
[/B]

  • I have to find the constant of normalisation for rho.

Homework Equations

The Attempt at a Solution


[/B]
I wrote p(r)=p_0 * (e^-r/R)/r

I then wrote it as
p_0^2 * integral from -infinity to +infinity of (e^-(r/)^2)/r^2 = 1

i tried to change variable by making z = r/R but then i was left with

1/R*p_0^2* integral of e^(-z^2) * (z^-2) = 1

I know that the integral of e^(-z^2) is the square root of pi but I don't know how to evaluate this with the z^-2 as well.

Have I started this correctly at all?

Thanks
 
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You made mistake when calculating ##(e^{-r/R})^2##.
 
ElectricEel1 said:
p_0^2 * integral from -infinity to +infinity of (e^-(r/)^2)/r^2 = 1
What does this integral represent? Why are you squaring the charge density? Why should the integral equal 1?
 
TSny said:
What does this integral represent? Why are you squaring the charge density? Why should the integral equal 1?

Part of my problem I think is that I'm not 100% sure on why I'm doing it. In my quantum work before the wave function was squared and multiplied by its complex conjugate and summed over infinity because the probability of finding it there will be 1. (I think). So I've just tried to use the same kind of method for this problem. Thanks
 
TSny said:
What does this integral represent? Why are you squaring the charge density? Why should the integral equal 1?
You are right, I completely misunderstood that radial function to be a wavefunction, didn't pay attention that it's a charge distribution already.
 
So should I not have squared this at all?
 
What have you learned from your electrodynamics course about charge distribution? For example given total charge ##Q##, its density ##\rho(\mathbf r)##, and volume ##V##, what's the equation connecting these quantities?
 
Your attempt at a solution is incomplete. What did you use for dV in your integral?
 
blue_leaf77 said:
What have you learned from your electrodynamics course about charge distribution? For example given total charge ##Q##, its density ##\rho(\mathbf r)##, and volume ##V##, what's the equation connecting these quantities?

oh yeah

rho = total charge / volume right or (ze)/(4/3pi*R^3) ?

and rho in this problem also = constant * (e^(r/R))/r

I think I need a volume integral here somewhere but I've not quite reached setting it up yet
 
  • #10
ElectricEel1 said:
rho = total charge / volume right or (ze)/(4/3pi*R^3) ?
This is true only if the volume charge density is constant, i.e. independent of r. Here it is not because it is given to you as a function of r.
ElectricEel1 said:
I think I need a volume integral here somewhere but I've not quite reached setting it up yet
You do need a volume integral that looks like ## \int \rho (r) dV ##. You know what ## \rho (r) ## looks like. What about ## dV ##?
 
  • #11
or maybe that means that i should integrate
kuruman said:
This is true only if the volume charge density is constant, i.e. independent of r. Here it is not because it is given to you as a function of r.

You do need a volume integral that looks like ## \int \rho (r) dV ##. You know what ## \rho (r) ## looks like. What about ## dV ##?
dV would look like r^2 sin theta right? if i do this then doesn't that mean that the integral is equal to the total charge?
 
  • #12
ElectricEel1 said:
dV would look like r^2 sin theta right?
Not right. A volume element dV in spherical coordinates is more than just r2sinθ. What is the correct form?
ElectricEel1 said:
if i do this then doesn't that mean that the integral is equal to the total charge?
Only if you integrate over the correct limits of integration. What are these limits?
 
  • #13
kuruman said:
Not right. A volume element dV in spherical coordinates is more than just r2sinθ. What is the correct form?

Only if you integrate over the correct limits of integration. What are these limits?
r^2 sin theta dr dtheta dphi

limits for r theta and phi are 0 to infinity, 0 to pi, and 0 to 2pi, respectively. but why do I need to find total charge to find this constant of normalisation?

thanks for your patience and advice
 
  • #14
ElectricEel1 said:
but why do I need to find total charge to find this constant of normalisation?
The proton has charge e = 1.6×10-19 C. The integral ## \int \rho (r) dV ## with the charge density you have will not give you that value or units. You need to multiply the charge distribution by normalization constant N so that the integral indeed gives you that value. This is what normalization means. When you have your answer, you should do a dimensional analysis to verify that your normalized ## \rho (r) ## has dimensions of charge/volume.

I am not sure that the upper integration limit for r extends all the way to infinity. What is the model for the proton that you are using according to the statement of the problem? Is the proton modeled as extending all the way out to infinity?
 
  • #15
kuruman said:
The proton has charge e = 1.6×10-19 C. The integral ## \int \rho (r) dV ## with the charge density you have will not give you that value or units. You need to multiply the charge distribution by normalization constant N so that the integral indeed gives you that value. This is what normalization means. When you have your answer, you should do a dimensional analysis to verify that your normalized ## \rho (r) ## has dimensions of charge/volume.

I am not sure that the upper integration limit for r extends all the way to infinity. What is the model for the proton that you are using according to the statement of the problem? Is the proton modeled as extending all the way out to infinity?

at the bottom of the sheet it has a note saying that the charge distribution falls away as r approaches infinity and that this form is not very realistic but was chosen to simplify the maths so i assumed that meant that the limits were 0 and infinity. does that sound possible?
 
  • #16
ElectricEel1 said:
so i assumed that meant that the limits were 0 and infinity. does that sound possible?
Yes, that's what it meant. Do you understand how to proceed from here?
 
  • #17
kuruman said:
Yes, that's what it meant. Do you understand how to proceed from here?
from there had p(r)=N*(e^-r/R)/r so had the integral of
N*(e^-r/R)/r * r^2 sin theta dr dtheta dphi = 1 and evaluated with limits 0>infinity,0>pi,0>2pi to get N = 1/(4*pi*R^2)
 
  • #18
Why is the integral equal to 1? You are not normalizing a probability which is equal to 1 when you add all probabilities over all space. You want to say that when you add up all the little charge elements ρdV over all space you get the total charge of the proton. So what should the other side of the equation be?
 
  • #19
kuruman said:
Why is the integral equal to 1? You are not normalizing a probability which is equal to 1 when you add all probabilities over all space. You want to say that when you add up all the little charge elements ρdV over all space you get the total charge of the proton. So what should the other side of the equation be?
so is it just equal to z*e or Q? the total charge?
 
  • #20
What's z for a proton?
 
  • #21
kuruman said:
What's z for a proton?
1!
 
  • #22
ElectricEel1 said:
so is it just equal to z*e or Q? the total charge?
Correct. So, what is the RHS of the equation when you substitute 1 for z?
 
  • #23
so right hand side is just e, or since the next question has 'q' in it i should probably use q
 
  • #24
Yes and yes.
 

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