# Uncertainty Principle & an Atomic Nucleus

## Homework Statement

A neutron in an atomic nucleus is bound to other neutrons and protons in the nucleus by the strong nuclear force when it comes within about 1 fm of another particle . What is the approximate kinetic energy in MeV of a neutron that is localised to within such a region? Take delta(x)*delta(p) = h/(4pi) and rest energy of neutron to be 939 MeV. Give your result to 2 significant figures.

## Homework Equations

$$\Delta$$X$$\Delta$$P = h/4$$\pi$$
KE = P2/2m

## The Attempt at a Solution

We know $$\Delta$$X = 1x10-15m and we know E0 = 939MeV and $$\Delta$$P can be easily determined. However, I don't know how to continue because I don't think we're given enough information. A couple people in my class have mentioned that we should assume $$\Delta$$P ~ P in which this would make the question simple. I don't understand this since our $$\Delta$$X is very small so our $$\Delta$$P would be very large. But the uncertainty in this case is just a number so it could be anything so how is it at all related to P hence why the hell can we just assume $$\Delta$$P ~ P?? Or are my classmates wrong like I think they are Related Introductory Physics Homework Help News on Phys.org
ojs
Well, the planck constant is of the order 10-34 so the $\Delta$P is of the order 10-19 which is not large in my book.

As to why this assumption is to be made my best guess would be that within a nucleus the particles are considered still in respect to one another, just a small vibration whose magnitude is given by the uncertainty principle with the lower bound being 0 (and so the upper bound is the magnitude of the uncertainty).

Well, the planck constant is of the order 10-34 so the $\Delta$P is of the order 10-19 which is not large in my book.

As to why this assumption is to be made my best guess would be that within a nucleus the particles are considered still in respect to one another, just a small vibration whose magnitude is given by the uncertainty principle with the lower bound being 0 (and so the upper bound is the magnitude of the uncertainty).
I don't completely understand but if you're saying that assuming $\Delta$P ~ P is okay then that's good enough for me. We've been getting heaps of these dodgy questions lately and they're starting to annoy me lol. Thanks.