Find out kinetic Energy using Rutherford Formula

[Juli]

Homework Statement

A narrow bundle of protons of uniform energy hits vertically on a 4µm
thick gold foil. The fraction $\eta$=1.35. 10-3 of the impacting protons are distributed around the angle $\theta$=60° scattered in the angle interval $d\theta$.
What kinetic energy do the incident protons have?
$\omega$

Relevant Equations

Rutherford's diffraction equation
$$ \frac{dn}{nd\Omega} = \frac{Z^2e^4DN}{(16\pi \epsilon_0)^2E_0 \sin^4(\frac{\theta}{2})}$$ with
$$ d\Omega = \sin\theta d\theta d\phi $$

Hello everyone,

I am working on this problem and I think I almost solved it, but then I noticed, that I do not know what values I have for dn, n and dθ.
Can anyone help me with this?

 

[Mark44]

First off, per forum rules, you must show what you have done.
Second, the equation you show is not well-formed LaTeX, so does not render correctly. It's complicated enough (and I could not find a link to this equation) that I don't feel confident in trying to correct it.

Third, your equation that follows "with" looks to be incorrect, as well. Is this what you were trying to write?
$$d\Omega = sin\theta \frac{d\theta}{d\phi}$$
If so, I don't understand what it means. There's a differential on the left side and a derivative on the right side. If the left side is a differential, the right side should be, as well.

but then I noticed, that I do not know what values I have for dn, n and dθ.

As a guess, dn is the differential of n (i.e., a small change in the number of particles), n is the number of particles, and dθ is a small change in the angle.

 

[Juli]

First off, per forum rules, you must show what you have done.
Second, the equation you show is not well-formed LaTeX, so does not render correctly. It's complicated enough (and I could not find a link to this equation) that I don't feel confident in trying to correct it.

Third, your equation that follows "with" looks to be incorrect, as well. Is this what you were trying to write?
$$d\Omega = sin\theta \frac{d\theta}{d\phi}$$
If so, I don't understand what it means. There's a differential on the left side and a derivative on the right side. If the left side is a differential, the right side should be, as well.
As a guess, dn is the differential of n (i.e., a small change in the number of particles), n is the number of particles, and dθ is a small change in the angle.

Thanks for your reply, I was working on the rendering the whole time, because I see, that it doesn't work. But I know that my Latex code is correct. I checked it in an editor.
Maybe I can upload pictures instead.

dn is the Number of particles scattered at the angle $d\Omega$
n is the number of particles that were shot at the gold foil

Like i said, I could do the rest of the problem, but obviously I have to figure out these two parameters with the help of the information given in the text to solve the problem, since I have so solve it for the kinetic energy.

 

[Juli]

Thanks for your reply, I was working on the rendering the whole time, because I see, that it doesn't work. But I know that my Latex code is correct. I checked it in an editor.
Maybe I can upload pictures instead.

dn is the Number of particles scattered at the angle $d\Omega$
n is the number of particles that were shot at the gold foil

Like i said, I could do the rest of the problem, but obviously I have to figure out these two parameters with the help of the information given in the text to solve the problem, since I have so solve it for the kinetic energy.

I'll try again:
$\frac{dn}{nd\Omega} = \frac{Z^2e^4DN}{(16\pi \epsilon_0)^2E_0 sin^4(\frac{\theta}{2})}$

 

[Juli]

with $d\Omega = sin\theta d\theta d\phi$

 

[Juli]

But thee formula is just for the understanding of the problem and to know what I'm talking about. I am just wondering how I am supposed to know how many incident particles there where. Is there a general amout that you would just assume? Because the text just says "bundle of protons". And these "bundle of protons" equals n.

 

[haruspex]

I do not know what values I have for dn, n and dθ

You do not need dn and n separately, just the fraction dn/n. What is the relationship between that and η?
dθ seems to be a given, so can remain in the answer if necessary, but I expect it will cancel.

 

[TSny]

In order to get a numerical value for the energy of the protons, I think you need to know the numerical value of $d \theta$. Hope I'm not overlooking something.

$d \Omega$ will include all possible values of $\phi$.

 

[Juli]

I think I got it now. Thanks to both of you. I think by definition $\eta$ equals $\frac{dn}{nd\theta}$ which you will get after integrating over $d\phi$.
Sometimes I find it a bit hard to understand which values are meant by the describing texts, which is obviously is a big part of the problem solving.

 

[Juli]

Hello everyone,
I was too confidet, I thought i figured it out. But again, my solution massively differs from the books solution and I checked everything so many times and can't figure out where my mistake is.
I tried to write down what I did understandably. I'm greatful for anyone who take the time to look over it!

 

[TSny]

I don't understand why you integrated over $\phi$ the way you did.

The picture I have in mind is this

The total solid angle subtended by the yellow ring would be the integration of $d \Omega = \sin \theta d\theta d\phi$ over $\phi$.

If you give me the numerical answer that you are supposed to get, then maybe I can work back to get the intended interpretation of the wording of the problem.

 

[Mark44]

I think I got it now. Thanks to both of you. I think by definition $\eta$ equals $\frac{dn}{nd\theta}$ which you will get after integrating over $d\phi$.

The definition above would be better, IMO, if it were written like this: $\eta = \frac 1 n \frac{dn} {d\theta}$. I'm assuming that the last part is the derivative of n with respect to $\theta$. Under normal circumstances derivatives like this shouldn't be treated as fractions although this can be done when one is solving differential equations.

 

[Juli]

Hey, thank you for your reply. I think I found out that the book made a mistake and my solution is right. If you have further interest in discussing the solution, let me know. Otherwise, thanks a lot again, I'm good now :)

 

[haruspex]

Your solution makes no sense to me.
As @TSny writes, $d\Omega$ is the slice of the spherical surface in yellow in post #2. I.e., $=2\pi\sin(\theta)d\theta$. No other integration is necessary.

Seems to me that the question cannot be solved numerically without a value for $d\theta$, as @TSny pointed out in post #8 in the earlier thread.

Btw, please do not create multiple threads for the same problem.

 

[berkeman]

Btw, please do not create multiple threads for the same problem.

@Juli -- What is different about this thread compared to your previous thread? https://www.physicsforums.com/threads/find-out-kinetic-energy-using-rutherford-formula.1058642/

Should I merge your two threads?

 

[Juli]

@Juli -- What is different about this thread compared to your previous thread? https://www.physicsforums.com/threads/find-out-kinetic-energy-using-rutherford-formula.1058642/

Should I merge your two threads?

Yes you can do that. Thank you. I initially wanted to make it less confusing, but as it seems it is sensful to put it back together.

 

[berkeman]

Done.