Question

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.

Answer 1

Answer:

      $K = \frac{h'}{8 m \ \Delta x^2}$K

Explanation:

The Heisenberg uncertainty principle is

          Δx Δp ≥ h' / 2

          h’ =$\frac{h}{2\pi }$

The kinetic energy of a particle is

          K = ½ m v²

           p = mv

           v = $\frac{p}{m}$

substitute

           K = $\frac{1}{2} \frac{p^2}{m}$

from the uncertainty principle,

           Δp = $\frac{h'}{2 \ \Delta x}$

we substitute

          K = $\frac{1}{2m} ( \frac{h'}{2 \ \Delta x})^2$

          $K = \frac{h'}{8 m \ \Delta x^2}$