Minimum kinetic energy of an Alpha particle confined to a nucleus

[Brianrofl]

Homework Statement

http://puu.sh/bz0km/092f2d8781.png

Homework Equations

h^2/2ma^2

The Attempt at a Solution

I've found the uncertainty of the kinetic energy through many means, and I'm confident that it's .052MeV.

The easiest method that gave me this answer was:

using http://puu.sh/bz0Fm/f45f952a89.png

((1.05*10^-34)^2 / 2(6.64*10^-27)(10^-14)^2 ) / 1.6*10^-19 = .052MeV

So I know that the kinetic energy has an uncertainty of .052MeV and that its typical energy is 5MeV, so wouldn't the answer just be 5MeV - .052MeV = 4.948MeV?

 

[TSny]

In this type of question, the uncertainty in momentum is generally taken to be an estimate of the minimum momentum itself. Thus, the KE of .052 MeV would represent an estimate of the minimum KE of the alpha particle inside the nucleus. Did you try using .052 MeV as the answer?

You also have to be careful in that different people will use somewhat different expressions for the right side of the uncertainty principle: h, h-bar, h-bar/2, etc.

 

[Brianrofl]

Wow, when I first put in my answer I put in .51 instead of .519, so it got marked as wrong :( I was right the whole time.

There is this problem, though -- http://puu.sh/bz4AT/4545a75828.png

I can't even find anything like it in my textbook, any tips just to get me started on it?

Thanks.

 

[TSny]

Wow, when I first put in my answer I put in .51 instead of .519, so it got marked as wrong :( I was right the whole time.

Is that meant to be 0.0519 MeV?

There is this problem, though -- http://puu.sh/bz4AT/4545a75828.png

I can't even find anything like it in my textbook, any tips just to get me started on it?

Thanks.

Are you familiar with the energy-time uncertainty principle? See http://pdg.web.cern.ch/pdg/cpep/unc_vir.html

 

[Brianrofl]

Is that meant to be 0.0519 MeV?

Yeah



Are you familiar with the energy-time uncertainty principle? See http://pdg.web.cern.ch/pdg/cpep/unc_vir.html

Ok, yeah I've used that equation before and I got the correct answer, thanks!