# Simple Measurement of Atomic Nuclei Radii

1. Mar 16, 2004

### Claire84

Okay, we have a graph that shows how v.high energy electrons are diffracted by atomic nuclei and the first minimum is shown to be at approximately 51 degrees for carbon-12 with energy 420MeV. To calculate the nuclear radii I decided to use sin(theta)= 1.22lambda/d and rearranged this so d was the subject of the eq. To work out lamba I used lamba= h/sqrt(2mE). Then to finally calculate the radius of this nucelus, I used R=d/2. However, I've ended up with an answer x10^-14 (and almost x10^-13).

Can someone please tell me what's wrong with this? I've used m as the mass of an electron and put the energy into joules so I've got 420x10^6x1.6x10^-19 J

2. Mar 17, 2004

### Claire84

I've checked my textbook and it says I'm using the right formulae so I really don't understand why I'm not getting my answer to x10^-15. Can anyone spot an error in what I've written above? Thanks.

3. Mar 21, 2004

### Claire84

How can you use Rayleigh's criterion here with such a large angle? I thought it could only be used for tiny resolving angles?

4. Mar 21, 2004

### Orion1

Beta Scattering...

Electron Wavelength:
$$\lambda_e = \frac{h}{ \sqrt{2m_eE}}$$

Scattering Criterion:
$$d \sin \Theta = m \lambda$$

$$d = \frac{mh}{ \sin \Theta \sqrt{2m_eE}}$$

$$d = 2r_n$$

$$r_n = \frac{mh}{ 2 \sin \Theta \sqrt{2m_eE}}$$

First minimum intensity:
$$m = \frac{1}{2}$$

$$r_n = \frac{h}{ 4 \sin \Theta \sqrt{2m_eE}}$$

rn = 1.925*10^-14 m

$$\lambda_e < d$$

$$\lambda_e$$ = 5.984*10^-14 m
d = 7.7*10^-14 m

Claire84, these equasions cannot resolve the radius of a nucleus unless the electron wavelength here ($$\lambda_e$$ = 5.984*10^-14 m) is less than the diameter of a nucleus, here dn = 5.5*10^-15 m.

Electron nuclear resolution criterion:
$$\lambda_e \leq d_n$$

Orion1-Claire Criterion:
$$\sin \Theta \sqrt{E} = \frac{mh}{ 2 r_0 \sqrt[3]{A_n} \sqrt{2m_e}}$$
r0 = 1.2*10^-15 m
An - target mass (amu)

Last edited: Mar 21, 2004