Phy5645/Uncertainty Relations Problem 2: Difference between revisions

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(New page: This 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.''' '''Determine the magnitude of ...)
 
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This a problem from the book of W.Greiner named Quantum Mechanics.  
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>


'''Consider there is a box with a particle (a nucleon) in it. Width of the box is L.'''
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>
'''Determine the magnitude of kinetic energy of the particle.'''




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


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.


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>


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>
<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}\,\text{cm})^2}  \approx 0.2\,\text{MeV}</math>


 
Back to [[Heisenberg Uncertainty Principle#Problems|Heisenberg Uncertainty Principle]]
Let the side of the box shrink <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.
 
 
<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>

Latest revision as of 13:23, 18 January 2014

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.


Back to Heisenberg Uncertainty Principle

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