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| This is 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> |
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| '''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.'''
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| [[Image:Untitled.gif|thumb|425px|a picture of the box taken from the same book]]
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| | Let the length of a side of the box <math>l= \Delta x \rightarrow 0 </math> |
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| 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>
| | 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. |
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| 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>
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| | <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> |
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| Let the side of the box shrink <math>L= \Delta x \rightarrow \ 0 </math>
| | Back to [[Heisenberg Uncertainty Principle#Problems|Heisenberg Uncertainty Principle]] |
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| 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.
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| <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>
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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