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An electron, tial well may be anywhere within the interval 2a. So the uncertainty in its position is Δx= 2a. There must be a corresponding uncertainty in the momentum of the electron and hence it must have a certain kinetic energy. Calculate this energy from the uncertainty relationship and compare it.

Respuesta :

Answer:

      [tex]K = \frac{h'}{8 m \ \Delta x^2}[/tex]K

Explanation:

The Heisenberg uncertainty principle is

          Δx Δp ≥ h' / 2

          h’ =[tex]\frac{h}{2\pi }[/tex]

The kinetic energy of a particle is

          K = ½ m v²

           p = mv

           v = [tex]\frac{p}{m}[/tex]

substitute

           K = [tex]\frac{1}{2} \frac{p^2}{m}[/tex]

from the uncertainty principle,

           Δp = [tex]\frac{h'}{2 \ \Delta x}[/tex]

we substitute

          K = [tex]\frac{1}{2m} ( \frac{h'}{2 \ \Delta x})^2[/tex]

          [tex]K = \frac{h'}{8 m \ \Delta x^2}[/tex]