Phy5645/Uncertainty Relations Problem 2: Difference between revisions

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This is a problem from the book of W.Greiner named Quantum Mechanics.  
This is a problem from the book of W.Greiner named Quantum Mechanics.  


'''Consider there is a box with a particle (a nucleon) in it. Width of the box is L.'''
'''Consider a box with a particle (a nucleon) in it. The width of the box is <math>l</math>. Determine the magnitude of kinetic energy of the particle.'''  
'''Determine the magnitude of kinetic energy of the particle.'''  


[[Image:Untitled.gif|thumb|425px|a picture of the box taken from the same book]]  
[[Image:Untitled.gif|thumb|425px|a picture of the box taken from the same book]]  




According to Heisenberg uncertanity principle: <math>\Delta p \Delta x \cong \hbar</math> and so <math> \Delta p\cong \frac{\hbar}{\Delta x}</math>. On the other hand, as we know that <math>E=\frac{\ P^2}{2m}</math>. Therefore,<math>\Delta E= \frac{(\Delta P) ^2}{2m}</math>
According to the Heisenberg Uncertanity Principle, <math>\Delta x\,\Delta p \cong \hbar</math> and so <math> \Delta p\cong \frac{\hbar}{\Delta x}</math>. On the other hand, as we know that <math>E=\frac{p^2}{2m}.</math> Therefore,<math>\Delta E= \frac{(\Delta p)^2}{2m}.</math>


If we plug <math> \Delta p </math> into the energy equation; it will look  <math> \Delta E\cong \frac{\hbar ^2}{2m(\Delta x) ^2}</math>
If we plug <math>\Delta p</math> into the energy equation, we obtain <math> \Delta E\cong \frac{\hbar ^2}{2m(\Delta x) ^2}</math>




Let the side of the box shrink <math>L= \Delta x \rightarrow \ 0 </math>  
Let the length of a side of the box <math>l= \Delta x \rightarrow 0 </math>  


Knowing that Nucleons have size of <math> \approx 10 ^{12}cm </math>, Nucleon mass <math> mc^2 \cong 938 MeV</math>, '''and''' <math>\hbar c\cong 197 \times 10 ^{-13} cm Mev</math>, then we can calculate kinetic energy.  
Knowing that the size of a nucleon is about <math>10^{12}\,\text{cm},</math> that their mass <math> mc^2 \cong 938\,\text{MeV}</math>, and that <math>\hbar c\cong 197 \times 10 ^{-13}\,\text{MeV}\cdot\text{cm}</math>, we can calculate kinetic energy.  




<math> \Delta E\cong \frac{\hbar ^2}{2m(\Delta x) ^2}= \frac{\hbar ^2 c ^2}{2mc ^2(\Delta x) ^2 }=  \frac{(197 \times 10 ^{-13} cm Mev) ^2} {(2) (938 MeV) (10 ^{-12}) ^2}  \approx 0.2MeV</math>
<math> \Delta E\cong \frac{\hbar ^2}{2m(\Delta x) ^2}= \frac{\hbar ^2 c ^2}{2mc ^2(\Delta x) ^2 }=  \frac{(197 \times 10 ^{-13}\,\text{MeV}\cdot\text{cm}) ^2} {(2) (938\,\text{MeV}) (10 ^{-12}) ^2}  \approx 0.2\,\text{MeV}</math>

Revision as of 12:31, 5 April 2013

This is a problem from the book of W.Greiner named Quantum Mechanics.

Consider a box with a particle (a nucleon) in it. The width of the box is . Determine the magnitude of kinetic energy of the particle.

a picture of the box taken from the same book


According to the Heisenberg Uncertanity Principle, and so . On the other hand, as we know that Therefore,

If we plug into the energy equation, we obtain


Let the length of a side of the box

Knowing that the size of a nucleon is about that their mass , and that , we can calculate kinetic energy.