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Neutron, minimum potential energy, heisenberg uncertainty

  • Thread starter Yroyathon
  • Start date
  • #1
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hi folks, last problem I can't get done here. I usually have a good idea of where to start, what to use. but i'm a bit flummoxed.

Homework Statement


A neutron, of mass m = 1.67 10-27 kg, is localized inside a carbon nucleus, of radius 7.0 fm (1 fm 10-15 m). Use the uncertainty relations to calculate a minimum (negative) potential energy for neutrons inside heavy nuclei.

Homework Equations


(deltax)*(delta p_x) > h-bar = h/(2*Pi), the uncertainty inequality


The Attempt at a Solution


i'm not sure how to calculate the potential energy of a particle.
I have an equation for the kinetic energy of a particle, K=p^2/(2*m), in terms of the momentum and mass.
the textbook gives an example of potential energy of a spring, but this problem doesn't involve springs soo...

could the other uncertainty inequality be used here? I refer to
(delta E) * (delta t) > h-bar

I'm not sure if this energy would be potential, kinetic, or total energy. and I also don't know how to involve time, so I could get rid of the delta t.

any tips would be great. my approach to write down all relevant equations and then try to piece them together with the given constants has not worked this time.

Thanks much,
Yroyathon
 
Last edited:

Answers and Replies

  • #2
79
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You are almost there. :)
So by using uncertainty principle, you can find the minimum KE for the neutron to be in Nucleus, right?
So this Neutron would have some positive KE, and therefore would actually move around.
But by in the real world, Neutron doesn't move around, much (and thanks that it doesn't!).
So obviously there must have some potential well, or negative potential energy to hold the neutron in place, so that its net energy is 0 (equilibrium. Basically not moving.)
So what is your potential energy?
 
  • #3
42
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ok, so I start with (deltax)/(delta p_x) > h/(2*Pi).
deltax = 2 * given radius.

so (delta p_x) > h/(2*Pi*2*r) . now I square both sides
(delta p_x)^2 > h^2/(4*Pi^2*4*r^2) , then divide by twice the mass

(delta p_x)^2 / (2*m) > h^2/(4*Pi^2*4*r^2*2*m) , and this left term is the KE

KE > h^2/(4*Pi^2*4*r^2*2*m)

is this right?

this is positive, so the negative of this will keep the total energy 0. this makes sense, i guess i didn't know about other forces contributing to the potential energy. but just considering the kinetic energy as the force that can cancel the potential energy is pretty helpful.
 
  • #4
42
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got it!

thanks so much for your help!
 
  • #5
Redbelly98
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Just a minor point (it won't change your answer):

The total energy should be ≤0, not necessarily =0, for a bound particle. Hence the wording of the question, to find the minimum potential.
 

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