Where does the first minimum appear in a double slit experiment with neutrons?

[erok81]

Homework Statement

In a double slit experiment, a beam of neutrons with speed 0.6m/s is diffracted by two apertures which are 0.05μm apart.

a. What is the neutron wavelength?
b. Where does the first minimum appear?

Homework Equations

\lambda = \frac{h}{p}

dsin\theta =(m+\frac{1}{2})\lambda

The Attempt at a Solution

First off for the wavelength.

\lambda = \frac{h}{p}

I had to calculate p first. Which is just mv -> (1.657x10^-27)(0.6) = 9.942x^-28

Plugging that into the de Broglie wavelength forumla...

\lambda = \frac{h}{p} = \frac{6.63\times10^{-34}}{1.005\times10^{-27}} = 6.6\times10^{-7}

That answers part a.

Now part b I am stuck on. First I don't know how far away the detector is, so I can't really give a definite location. So I was just going to go with the angle.

Using dsin\theta =(m+\frac{1}{2})\lambda with m= 0

I get...

sin^{-1}\left[\frac{1}{2}\times\frac{6.6\times10^{-7}}{0.05\times10^{-6}}\right] = sin^{-1}(sin{\theta})

Before taking the arcsin I have a value of 6.597. Which obvisouly gives me a weird answer for theta. Weird answer being 1.571° and -2.574°.

Any ideas where I am going wrong?

 

[vela]

I think your wavelength is off by a factor of 1000.

Well, maybe not. Let me recheck my calculations.

 

[erok81]

I hope it is. That would give me a better answer. I rechecked and I had p wrong in the post, which I've corrected, but in my wavelength calculation I had it right. So lamda is correct...at least for me.

 

[vela]

Your answer is correct. My mass was off by a factor of 1000. So the equation is telling you there is no minimum.

I'm wondering, though, if your velocity is correct; 0.6 m/s is pretty slow. Are you sure it's not supposed to be v/c=0.6, where c is the speed of light? Of course, if that's the case, the wavelength comes out incredibly small.

 

[erok81]

Your answer is correct. My mass was off by a factor of 1000. So the equation is telling you there is no minimum.

I'm wondering, though, if your velocity is correct; 0.6 m/s is pretty slow. Are you sure it's not supposed to be v/c=0.6, where c is the speed of light? Of course, if that's the case, the wavelength comes out incredibly small.

That's what I thought as well so I emailed the TA. He said it was right.

But like you said, maybe there is no minimum. Since without knowing the distance to the detector I can't really give an answer, only the angle.

 

[erok81]

Although...it's a fairly popular method of using neutrons and double slit experiments. At least from google. So if there is no minimum, I wonder what kind of info can be gained from this type of experiment.

 

[vela]

Probably not much, other than to place a lower bound on the distance between the scattering centers.

 

[erok81]

Perfect. Thanks for the help. Good to know I didn't mess it up. :)