Phy5645/Uncertainty Relations Problem 2

According to the Heisenberg Uncertanity Principle, $\Delta x\,\Delta p\cong \hbar$ and so $\Delta p\cong {\frac {\hbar }{\Delta x}}$. On the other hand, as we know that $E={\frac {p^{2}}{2m}}.$ Therefore,$\Delta E={\frac {(\Delta p)^{2}}{2m}}.$

If we plug $\Delta p$ into the energy equation, we obtain $\Delta E\cong {\frac {\hbar ^{2}}{2m(\Delta x)^{2}}}$

Let the length of a side of the box $l=\Delta x\rightarrow 0$

Knowing that the size of a nucleon is about $10^{12}\,{\text{cm}},$ that their mass $mc^{2}\cong 938\,{\text{MeV}}$, and that $\hbar c\cong 197\times 10^{-13}\,{\text{MeV}}\cdot {\text{cm}}$, we can calculate kinetic energy.

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

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