Homework Help: A particle confined to a space

1. Apr 1, 2012

hover

1. The problem statement, all variables and given/known data
Neutrons and protons in atomic nuclei are confined within a region whose diameter is about
10^-15m = 1 fm

a) At any given instant, how fast might an individual proton or neutron be moving?
b) What is the approximate kinetic energy of a neutron that is localized to within such a region?
c) What would be the corresponding energy of an electron localized to within such a region?

2. Relevant equations

uncertainty principle
ΔxΔp ≥ h/(4π)

3. The attempt at a solution

I am having trouble visualizing how to go about the problem. I know I need to use the certainty prinicple in some way but I don't know how. I don't know if this is the right way to start but since we are confined to a sphere, would it be proper to apply the uncertainty principle in 3 directions?
Δx*mΔv ≥ h/(4π)
Δy*mΔv ≥ h/(4π)
Δz*mΔv ≥ h/(4π)
Solve for Δv for each direction and then use them to find the total magnitude.

If I can understand this then all the parts to this question will be simple.

2. Apr 1, 2012

Dick

Approximations using the uncertainty principle are just that. Approximations. I would think you would be ok with doing it in one direction. That should give you a good enough estimate.

3. Apr 2, 2012

hover

Ok so I will do everything in one dimension. Working with a proton(the work would be similar for a neutron)

a.)
Δx*mΔv ≥ h/(4π)
Δv ≥ h/(4π*m*Δx)
Δv ≥ 6.626E-34/(4π*1.67E-27*1E-15)

So the minimum speed for a proton confined to that particular space will be
= 31573672.2 m/s or 31573.6722 km/s or 31.5736722 Mm/s

b.)
A proton(I know it asks for E of a neutron but if I can do this general calculation then a neutron won't be much different) will have a minimum kinetic energy of

E = (1/2)mv^2
E = (1/2)(1.67E-27)(31573672.2)^2 = 8.324E-13 joules of energy

c.)
Is the same stuff except I'm dealing with an electron.

Is what I have correct?

4. Apr 2, 2012

Dick

Yes, except you've only got 3 decimal places of accuracy in the proton mass and this is an approximate calculation anyway. So writing 31573672.2 m/s is kind of silly. Round it off. And is Mm/s really a unit?

5. Apr 4, 2012

hover

Ok I'll round the number a bit since this is an approximate calculation. By the way, what prevents Mm/s from being a unit?

6. Apr 4, 2012

Dick

You mean mega-meters/sec? I'm not saying it COULDN'T be a unit, but I've never seen it used. If you used it in a problem there's a pretty good chance it would be taken for millimeters/sec.

7. Apr 4, 2012

hover

Yeah mega-meters/sec is what I meant by that. I've never seen it used like this either. I was really just playing with SI prefixes just for the hell of it. It does look cleaner though:

31500 km/s or 31.5*10^3 km/s
vs
31.5 Mm/s

It could be taken as millimeters too but generally that is written as mm not Mm.

8. Apr 4, 2012

Dick

Sure, I'd just stick with commonly used units. I don't see much usage of mega-meters.