Heisenberg Uncertainty Principle homework

[Ecoi]

Well first off, I am confused about what the book says earlier and what the actual answers are in the back of the book on homework problems. I thought I understood the book, but it seems like I don't.

The book has:

"Consider a particle whose location is known within a width of L along the x axis. We then know the position of the particle to within a distance Δx =< L /2. The uncertainty principle specifies that Δp is limited by

Δp >= h / (4piΔx) >= h / (2 L pi)

Because p = mv, we have Δp = mΔv, and

Δv = Δp / m >= h / (2mL pi)...

Kmin = (pmin)^2 / 2m >= (Δp)^2 / 2m >= h^2 / (8m L^2 * pi^2)"

Here is where I am confused. It seems like later in the book, they have Δx =< L in order to do calculations. So my question is, is Δx the radius away from a point or is it the diameter? From the above, I assumed that Δx is less than or equal to the radius. But in the problem below, I did that and I was off by (1/4) on my answer...

Homework Statement

A neutron is confined in a deuterium nucleus (deuteron) of diameter 2 x 10^-15 m. Use the energy-level calculation of a one-dimensional box to calculate the neutron's minimum kinetic energy. What is the neutron's minimum kinetic energy according to the uncertainty principle?

Homework Equations

Equations

Δx Δp >= h / (4pi)

For a one dimensional box:

En = n^2 * h^2 / (8 m L^2) where L is the length of the box

Constants:

h = 6.6261 x 10^-34 J * s = 4.1357 x 10^-15 eV * s

mass of neutron = 1.6749 x 10^-27 kg = 939.57 MeV / c^2

The Attempt at a Solution

For the one-dimensional box method I get:

E1 = h^2 / 8 m L^2 = 51.1 MeV

Okay, the book got 51.1 MeV, so I am happy with this. Now,

Δx Δp >= h / (4pi)

We know m and Δx

I took Δx = radius of atom and so I obtain:

Δp = 5.273 x 10^-20 kgm/s

So,

E = (Δp)^2 / 2m = 5.18 Mev

Wrong!

Book gets 1.30 Mev. If you divide my answer my 4 you get the book's answer (because of a difference of 1/2 in the Δp)

So ok, so Δx = L. Then what in the world was the book talking about above with Δx = L / 2? I guess I could just remember to do this in each calculation, but I am either misunderstanding something with the problem, or misunderstanding something with what the text explained. Am I being retarded or something and it's so obvious or what?

Also, I derived a formula with Δx = L and I obtained:

Kmin = h^2 / (32 m pi^2 L^2)

If you use this on the above problem, you get 1.30 MeV! The correct answer! Is the book incorrect or what happened?

*Edit: I was wondering, would I ever need to consider a relativistic case? If so, I wasn't exactly sure if I could simply have (ΔE)^2 = (Δp)^2 * c^2 + (E0)^2, but rather have to write it out, etc. Seems like it would be troublesome. I don't think we need to be able to do that in my class, but I am just curious and was wondering how that would work out if someone doesn't mind explaining that to me as well.

 

[_coyote_]

Yes, you may need to consider a relativistic case. In general, the uncertainty principle for relativistic particles is ΔEΔt ≥ h/(4π). This means that the uncertainty in energy and time are related, so if one of them is known, the other can be determined. In addition, the uncertainty in momentum and position are related, with ΔpΔx ≥ h/(4π). For a relativistic particle, the energy-momentum relation is E^2 = p^2*c^2 + (m_0*c^2)^2, where m_0 is the rest mass of the particle. This means that the uncertainty in energy can be determined from the uncertainty in momentum: ΔE^2 = (Δp)^2*c^2 + (E_0)^2. So in your case, you would have to use the uncertainty principle to determine Δp, and then use the energy-momentum relation to determine ΔE.