Calculating Neutron Peaks with Increased Kinetic Energy

[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


[tex]E=\frac{\hbar^{2}k^{2}}{2m}[\tex] where <br /> <br /> [tex]k=\frac{2\pi}{\lambda}[\tex] and <br /> <br /> $$\lambda=2d\sin\theta[\tex]<br /> <br /> <br /> I would appreciate any help on this. <br /> <br /> thanks <br /> <br /> newo$$[/tex][/tex]

 

[NEWO]

hmm my latex doesn't seem to work here, don't know why, can you get the gist of it?

 

[Hootenanny]

hmm my latex doesn't seem to work here, don't know why, can you get the gist of it?

You need to swap the back-slashes to forward slashes like this (without the spaces)

[ /tex ]

As for your question, how does this;

$$\lambda=2d\sin\theta$$

Relate to n? Remember than for the maximum n $\Rightarrow \sin\theta = 1$, also note that the number of peaks = n+1

 

[NEWO]

ahhh so

$$\lambda=nd\sin\theta$$

or am I missing something

 

[Hootenanny]

ahhh so

$$\lambda=nd\sin\theta$$

or am I missing something

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?

 

[NEWO]

yeah as the energy increases the number of peaks increases also

thanks for your help it is much appreciated

 

[Hootenanny]

yeah as the energy increases the number of peaks increases also

thanks for your help it is much appreciated

Sounds good to me. My pleasure