Quantum mechanics uncertainty principle

[sawhai]

Homework Statement

If a neutron is confined to the nucleus, its wave function
– and therefore the probability of finding it at a certain position – has a
characteristic width, Δ
x, equivalent to the nucleus diameter. What is the uncertainty
in the neutron momentum Δ
p as a function of atomic mass number
A? What energy would a neutron with momentum
p equivalent to the momentum uncertainty Δ
p correspond to (justify using a
non-relativistic approximation) in MeV?

Homework Equations

r=(n^2h^2ε0^2)/(πmeZe^2); ΔxΔp>h

The Attempt at a Solution

My guess is that in order to find an expression for Δp in terms of A I should use Bohr radius equation but I'm not sure how to go from there

 

[BruceW]

hmm. The Bohr radius is effectively the average radius of the electron from the nucleus, but you're trying to find the width of the nucleus. So I don't think that is the right equation.

To find the radius of the nucleus as a function of atomic mass number, you need to use a certain model of the nucleus which is often used in these kinds of calculations.

 

[sawhai]

Youa are right. I realized that Bohr radius won't take me anywhere. I used the radius equation R=r0*A^(1/3). I Plugged in this radius for Δx into ΔPΔx>=h/2 with r0=1.2*10^-15 and h=0.658*10^-15 and got an equation Δp=(82.27/A^(1/3)) MeV/c. Then I used KE=(pc)^2/(2*(m0c^2) using Δp I derived and rest mass of the neutron (939.56MeV) and got 3.6/A^(1/9).
Do you think I did this correct?

 

[BruceW]

It is all good except the power which A is raised to. It was 1/A^(1/3) for the momentum, which is correct. But then for the energy, the momentum was squared, so A should not be to this power: 1/A(1/9)

EDIT: sorry if I am not explaining very well. I mean that your answer was 3.6/A^(1/9) and the number 3.6 is correct, but the A should not be raised to that power. (It looks like you have squared the original power, but that is not what should happen).

 

[sawhai]

My bad. It should be 1/A^(2/3). Thanks.

 

[sawhai]

For the next part I am asked to find an electron energy which is confined within Δx. I used the same Heisenberg uncertainty and used relativistic energy expression E= sqrt{(pc)^2+m0^2c^4}and found the electron energy to be {82.27/A^(2/3) + 0.511} MeV.
Now I have to find the depth of the confining potential energy and I assume that I have to use the electrostatic potential energy equation U=q1q2/4piεr. Is this correct? And should I use charges of electron and proton for q1 and q2 and Δx for radius?
I have to calculate this potential energy and show that an electron can't be confined in nucleus. I wish if I had a number for r:)

 

[BruceW]

{82.27/A^(2/3) + 0.511} MeV I don't think this is right, you didn't square the numbers, and you haven't square-rooted the whole thing to get the energy.

The depth of the confining potential energy of the nucleus is a tricky question. You're right that the electron would feel the electromagnetic force (not the strong force). Since this is just an order-of-magnitude question, and I can't think of any other way, I'd say you're right to use U=q1q2/4piεr as the depth of the well, with Δx for the radius. But you wouldn't be using the charge of just one proton, since there are several inside the nucleus.

 

[sawhai]

I again made a mistake in my calculation. It should be sort {(82.7^2)/A^(2/3)+0.511^2}MeV (math has never been my favorite). And as you mentioned the electrons only interact electromagnetically with the nucleus. So we will have A-Z protons and A-Z electrons to consider plugging in the potential energy equation? And the way to proof that an electron can't be confined inside the nucleus is to show that the uncertainty in its energy is greater than the potential energy that is supposed to hold the electron inside the nucleus?
Many thanks

 

[BruceW]

Yep, that all looks good. But I would say just one electron, since the question is asking about confining an electron to the nucleus.