What Is the Width of the Region on the Detector Hit by Neutrons?

[Pickled_Gorilla]

Homework Statement

A collection of neutrons, nominally at rest, are confined in a region 1.0 nm wide on the x-axis at a height of 50 cm above a neutron detector. The neutrons are released and fall under the influence of gravity towards the detector which records the horizontal position of the hits. Estimate the width of the region of the detector that gets hit by neutrons.

Homework Equations

(1) E_{0 neutron} = 939.57 MeV
(2) E = p²/2m
(3) p = h/λ
(4) a sin(θ) = mλ, a is slit width (1 nm)
(5) y= Dmλ/a, y is distance on detector, D is distance to detector from slit (50 cm), and a is slit width (1 nm)

The Attempt at a Solution

I treated this problem as a single slit diffraction problem.
Since the neutrons are initially at rest, their energy is E₀.
Using (2) to find p, and then (3) to solve for λ gives 9.331 * 10^-16m.

Setting sin(θ) = 1 to find the maximal minima, (4) becomes m = a/λ, and solving for m gives 1.07*10⁶.
Plugging into (5) gives 0.5 meters, and since this is the distance in one half of the picture, the number is doubled for the total area. Thus the total area on the detector hit by neutrons is 1 meter.

Am I approaching this problem correctly?

 

[haruspex]

I treated this problem as a single slit diffraction problem.

Not sure whether that is valid. It sounds more a circular hole than a parallel-sided slit.

Setting sin(θ) = 1 to find the maximal minima

I'm sure that is not valid. That will only be a minimum if a/λ is an integer, and anyway the distance to it would be infinite.
Your eqn (5) is an approximation for small θ.
Taking the first minimum would seem more reasonable.

 

[Pickled_Gorilla]

I talked to my professor, It is actually a Heisenberg Uncertainty Principle problem. Δx is 0.5 nm, then you can solve for the Δp. use E = mgh to solve for the ΔE in the time related H.U.P. and then use the velocity from Δp multiplied by the time to get displacement. Finally double the displacement to get total width. I ended up getting 3.233 μm.

 

[haruspex]

I talked to my professor, It is actually a Heisenberg Uncertainty Principle problem. Δx is 0.5 nm, then you can solve for the Δp. use E = mgh to solve for the ΔE in the time related H.U.P. and then use the velocity from Δp multiplied by the time to get displacement. Finally double the displacement to get total width. I ended up getting 3.233 μm.

Ah, of course; that crucial adjective:

nominally at rest

For what it is worth, if I use your original method but my approach of first minimum I get 1μm.
But since the HUP method involves descent time, which rises as the square root of the distance, the similarity of the answers must be coincidence.