Setting up an Uncertainty Problem

AI Thread Summary
The discussion revolves around calculating the minimum uncertainty in a proton's velocity within a gold nucleus, which has a radius of approximately 6 femtometers (fm). The initial misunderstanding involved interpreting the 6 fm as a diameter instead of recognizing it as a radius, leading to confusion about the uncertainty in position (Δx). The correct approach is to consider the full range of uncertainty as the entire diameter of 12 fm, not just half. This clarification emphasizes the importance of accurately defining the parameters in quantum mechanics problems. Understanding these concepts is crucial for correctly applying the uncertainty principle in calculations.
member 545369

Homework Statement


A proton is known to lie within the nucleus of a gold atom. The radius of a gold nucleus is approximately 6 fm. What is the minimum uncertainty in the proton’s velocity [you may treat the problem as one-dimensional and you should express your answer as a fraction of c].

Homework Equations


ΔxΔp ≥ ħ/2

The Attempt at a Solution


I think I got this, but Chegg is showing up weird inconsistent answers. I just want to make sure my logic is sound:

So since we can treat this one dimensionally, we know that the proton lies within a 6fm range. For simplicity, we can (in our imaginations) draw a 6 fm line and put a point on the center. On that center, our uncertainty of the position of the proton is ± 3 fm. So our Δx should be 3fm instead of 6 fm!

The rest of the work is rather simple, I just want to make sure I'm setting this up properly.
 
Physics news on Phys.org
The 6 fm represents the radius of the nucleus.
 
TSny said:
The 6 fm represents the radius of the nucleus.

Not very helpful. I think my work shows that I understand this.
 
talrefae said:
Not very helpful. I think my work shows that I understand this.
No, it shows you thought the 6fm was a diameter.
 
  • Like
Likes PeroK
Not that it would make a difference to this order of magnitude calculation, but if someone gave me a proton in a one-dimensional box that extends, say, from zero to 12 fm and asked "where is the proton?", I would say "somewhere between zero and 12 fm". So in such situations, I consider the position uncertainty to be the entire range in which the particle can be without me knowing any better.
 
haruspex said:
No, it shows you thought the 6fm was a diameter.

You're 100% right. Thanks for that!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top