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 Assuming an electron is confined to a 1nm wide region, find the uncertainty in momentum using the Heisenberg Uncertainty principle. You can assume the uncertainty in position \Delta X as 1nm. Assuming p=\Delta p, find the energy of the electron in electron volts.

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As the electrons rotate in a circular path,

\Delta r=1nm=10^{-9}m

  \Delta p=\frac{h}{\Delta X}\\ \Delta p=\left ( \frac{331}{314} \right )*10^{25}\\ E=\frac{1}{2}mv^{2}=\frac{\Delta p^{2}}{2m}\\ E=3.8*10^{-2}eV

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