• Support PF! Buy your school textbooks, materials and every day products Here!

Particle in a box. KE from Δx?

  • Thread starter Hertz
  • Start date
  • #1
180
8

Homework Statement



Suppose a particle is confined in one dimension to a region of width L. Obtain an approximate formula for its minimum kinetic energy.

Homework Equations



ΔxΔp ≥ h/4π

The Attempt at a Solution



1. Put L in for Δx
2. Divide by L. It is positive
3. Δp ≥ h/(4πL)
4. KE = (p^2)/2m

Now what? Ok, so you're telling me that the uncertainty in the momentum can be determined. What does that say about the actual value of the momentum? Wouldn't the minimum momentum technically be zero? Wouldn't this mean the minimum KE is zero? There is no way this is right... It would be too easy..

Please please please help with this. My teacher is so bad and this of all classes is the class I'm falling behind in.. It's his first semester teaching and he just runs through calculations all day and apparently expects us to learn all this stuff at home.. I don't have time for this. A simple explanation will do me wonders. Thanks so much
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618

Homework Statement



Suppose a particle is confined in one dimension to a region of width L. Obtain an approximate formula for its minimum kinetic energy.

Homework Equations



ΔxΔp ≥ h/4π

The Attempt at a Solution



1. Put L in for Δx
2. Divide by L. It is positive
3. Δp ≥ h/(4πL)
4. KE = (p^2)/2m

Now what? Ok, so you're telling me that the uncertainty in the momentum can be determined. What does that say about the actual value of the momentum? Wouldn't the minimum momentum technically be zero? Wouldn't this mean the minimum KE is zero? There is no way this is right... It would be too easy..

Please please please help with this. My teacher is so bad and this of all classes is the class I'm falling behind in.. It's his first semester teaching and he just runs through calculations all day and apparently expects us to learn all this stuff at home.. I don't have time for this. A simple explanation will do me wonders. Thanks so much
In quantum mechanics you can't know an exact value for position and momentum simultaneously. If you knew the momentum exactly, you would know nothing about position. In this case, you know an approximate position, which says you can only know an approximate momentum. Neither is definite. The details involve knowing a position distribution is related to a momentum distribution by a fourier transform. They are both wave function distributions, not exact values. The uncertainty relation gives you a best case for how what you know about position is related to what you know about momentum. In that sense, it's a minimum.
 
  • Like
Likes 1 person
  • #3
38
1
I wouldn't use the uncertainty principle, as the above poster mentioned, it's all about the error bars that go on the measurement.

Consider that the particle is a standing wave in the box. It's energy is related to its wavelength. If you want to know the minimum energy you need to know the longest wave you can fit in the box. Well, the wave function has to be zero at the endpoints, so you can fit a half wavelength in the box, so [itex]\lambda=2L[/itex].

The kinetic energy of a particle is given by [itex]E=\frac{p^2}{2m}=\frac{h^2}{2m\lambda^2}[/itex] (Remember that [itex]p=h/\lambda[/itex])

I've given you the wavelength for the particle with minimum energy and the energy as a function of the wavelength, mash em together.
 

Related Threads on Particle in a box. KE from Δx?

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
968
Top