# Neutron scattering

1. Jun 8, 2006

### NEWO

hi,

in neutron scattering, if the lowest kinetic energy of a neutron is increased by a factor of 2, how do you work out the number of peaks produced?

I have worked out the lowest kinetic energy for a beta-brass CuZn to be 2.37meV using

$$E=\frac{\hbar^{2}k^{2}}{2m}[\tex] where [tex]k=\frac{2\pi}{\lambda}[\tex] and [tex]\lambda=2d\sin\theta[\tex] I would appreciate any help on this. thanks newo 2. Jun 8, 2006 ### NEWO hmm my latex doesn't seem to work here, dunno why, can you get the gist of it??? 3. Jun 8, 2006 ### Hootenanny Staff Emeritus You need to swap the back-slashes to forward slashes like this (without the spaces) [ /tex ] As for your question, how does this; Relate to n? Remember than for the maximum n $\Rightarrow \sin\theta = 1$, also note that the number of peaks = n+1 Last edited: Jun 8, 2006 4. Jun 8, 2006 ### NEWO ahhh so [tex] \lambda=nd\sin\theta$$

or am I missing something

5. Jun 8, 2006

### Hootenanny

Staff Emeritus
Almost, I believe it is;

$$\lambda = \frac{d\sin\theta}{n} \Leftrightarrow n = \frac{d\sin\theta}{\lambda}$$

Can you see what happens if you increase the kinetic energy of the particle?

6. Jun 8, 2006

### NEWO

yeah as the energy increases the number of peaks increases also

thanks for your help it is much appreciated

7. Jun 8, 2006

### Hootenanny

Staff Emeritus
Sounds good to me. My pleasure