Deriving Momentum Component from Single-Slit Neutron Diffraction Pattern

[bon]

URGENT - Quantum help needed!

Homework Statement

A beam of neutrons with known momentum is diffracted by a single slit. SHow that an approximate value of the component of momentum of the neutrons in a direction perpendicular to both the slit and the incident beam can be derived from the single-slit diffraction pattern.

Homework Equations

The Attempt at a Solution

No idea how to do this really.. Why would the neutrons have a momenum component in that direction anyway?

 

[hikaru1221]

Think of wave-particle duality. The neutrons should exhibit both properties. Then when a wave passes through a single slit, what will happen?

 

[bon]

They will diffract i guess.. so measuring from the centre of the slit, how am i meant to work out the standard deviation delta x ?

Also how do we work out a value for the standard deviation in momentum? Also, surely the momentum of the neutrons is in the direction they are traveling in,rather than perpendicular to it?

Thanks

 

[hikaru1221]

Think of the observed pattern: http://en.wikipedia.org/wiki/File:Diffraction1.png
Because intensity ~ number of particles, plus that the central maximum has very high intensity compared to the others, we may assume that most neutrons fall onto the region of central maximum. And you can calculate the angle corresponding to that maximum, can't you? This, plus the momentum p, will give you the needed result.

P.S.: The result only shows the so-called "average" value of the component of momentum along that direction. That doesn't mean every neutrons must go along that direction.

 

[bon]

Argh. Still not sure what to do. Sorry. So should i just call the separation a, the wavelength lambda, then how do I work out delta x? What about delta p?

 

[hikaru1221]

Have a look at this: http://hyperphysics.phy-astr.gsu.edu/Hbase/phyopt/sinslit.html#c1
The angle position of the 1st order minimum: $$sin\theta _1= \lambda / a$$ . As the central maximum is bounded by the two 1st order minimums, most neutrons are deflected by angles ranging from 0 to $$\theta _1$$. So if we take $$\theta _1$$ as a typical value, and assume that the neutron is not affected by any external force field (so that its energy remains the same; so do its speed and its momentum's magnitude), the component of momentum in the mentioned direction is $$psin\theta_1$$ . The momentum $$p = h/\lambda$$ .
Therefore: $$p_x = h/a$$ .

As I said, this only gives you a so-called "average value" of component of momentum in that direction. Each neutron should have its own corresponding component. If you apply the Heisenberg Uncertainty principle, what you get is somewhat different:
_ The uncertainty of position = the width of the slit: $$\Delta x = a$$
_ From Heisenberg Uncertainty principle: $$\Delta x\Delta p_x\geq h/2\pi$$ (or maybe $$h/4\pi$$) , we have: $$\Delta p_x=h/2\pi a$$
_ We have: $$p_x = \Delta p_x / 2 = h/4\pi a$$ as $$\vec{p}_x$$ can be in either one of two opposite directions +x and -x.

The two results are different because they are both estimated values. Each neutron must has its own momentum's component along x direction. However both results agree on the order of magnitude of $$p_x$$ , which convinces us that both are valid estimation.

 

[bon]

Thanks so much!