Rutherford scattering - electromagnetic form factor

• skrat
In summary, The conversation discusses Rutherford scattering and the electromagnetic form function, which can be written as $$F(q)=2\pi\int _0 ^1 d(cos\Theta)\int _0 ^\infty r^2dr\rho(r)e^{iqrcos\Theta}.$$ The final step involves integrating over the entire nucleus, resulting in $$F(q)=\frac{4\pi }{q}\int \rho (r)re^{\frac{iqr}{2}}sin(\frac{qr}{2})dr,$$ which explains the presence of ##sin(qr)##.
skrat

Homework Statement

Hi there. This is not really a problem, I am only trying to understand something but I simply can't.
So Rutherford scattering says that $$\frac{d\sigma }{d\Omega}=(\frac{Ze^2m}{8\pi \varepsilon _0 p^2})^2\frac{1}{\sin ^4(\Theta/2)}|F(q)|^2$$ where $$F(q)=\int \rho (\vec x)e^{i\vec q \cdot \vec x}d^3 \vec x$$ is the electromagnetic form function. Now for spherical symmetry: $$\rho (\vec x) \rightarrow \rho(x)$$ and $$\vec q \cdot \vec x=qrcos\Theta.$$ Meaning that my form function can be written as $$F(q)=2\pi\int _0 ^1 d(cos\Theta)\int _0 ^\infty r^2dr\rho(r)e^{iqrcos\Theta}=$$ $$=\frac{4\pi }{q}\int _0 ^\infty \rho(r)sin(qr)rdr.$$

I don't understand that last step. How the hell did we get ##sin(qr)##...?

The Attempt at a Solution

Ok, tried to do it step by step in order to answer this question but this is as close as I could get:
Starting with
$$F(q)=2\pi\int _0 ^1 d(cos\Theta)\int _0 ^\infty r^2dr\rho(r)e^{iqrcos\Theta}=$$ $$=2\pi \int \rho (r)r^2dr\int e^{iqrcos\Theta}d(cos\Theta)=$$ $$=2\pi \int \rho (r)r^2dr \frac{1}{iqr}[e^{iqr}-1]=$$ $$=2\pi \int \rho (r)rdr \frac{1}{iq}e^{\frac{iqr}{2}}[e^{\frac{iqr}{2}}-e^{-\frac{iqr}{2}}]=$$ $$=\frac{4\pi }{q}\int \rho (r)re^{\frac{iqr}{2}}sin(\frac{qr}{2})dr$$

Now I wonder what am I doing wrong...?

Last edited:
Make sure you are using the correct limits for the integral ##d(\cos\Theta)##.

Aaaaaaaamm... Ah, of course, I have to integrate over the entire nucleus (sphere)...

So the ##d(\cos)## integral is actually from 1 to -1. And this brings me to ##sin(qr) / qr##.

Last edited:
Mind = blown

1. What is Rutherford scattering?

Rutherford scattering is a phenomenon that occurs when a beam of particles, usually alpha particles, is directed towards a target material. The particles interact with the nuclei of the target atoms, causing them to scatter in different directions.

2. How is Rutherford scattering related to the electromagnetic form factor?

The electromagnetic form factor is a mathematical expression that describes how the charge is distributed within a particle. In Rutherford scattering, the incoming particles are affected by the electromagnetic form factor of the target nuclei, causing them to scatter in different directions.

3. How did Rutherford's experiment contribute to our understanding of the electromagnetic form factor?

Rutherford's famous experiment involved directing alpha particles at a thin gold foil and observing their scattering patterns. By analyzing these patterns, Rutherford was able to determine that the majority of an atom's mass and positive charge is concentrated in its nucleus, providing evidence for the existence of the electromagnetic form factor.

4. What is the significance of the Rutherford scattering - electromagnetic form factor relationship in modern physics?

The relationship between Rutherford scattering and the electromagnetic form factor is important in understanding the structure of matter and the nature of subatomic particles. It has also been used in various experimental techniques, such as electron scattering, to study the properties of particles and their interactions.

5. Are there any practical applications of Rutherford scattering and the electromagnetic form factor?

Yes, there are various practical applications of Rutherford scattering and the electromagnetic form factor. For example, the principles of Rutherford scattering are used in medical imaging techniques, such as PET scans, to detect and diagnose diseases. The electromagnetic form factor is also used in particle accelerators to study the properties of subatomic particles and in nuclear physics to understand the behavior of nuclei.

• Introductory Physics Homework Help
Replies
17
Views
518
• Introductory Physics Homework Help
Replies
6
Views
226
• Introductory Physics Homework Help
Replies
5
Views
445
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
64
Views
2K
• Introductory Physics Homework Help
Replies
12
Views
259
• Introductory Physics Homework Help
Replies
4
Views
453
• Introductory Physics Homework Help
Replies
21
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
1K