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

`K = (h')/(8 m \ \Delta x^2)`K

Step-by-step explanation:

The Heisenberg uncertainty principle is

Δx Δp ≥ h' / 2

h’ =
`(h)/(2\pi )`

The kinetic energy of a particle is

K = ½ m v²

p = mv

v =
`(p)/(m)`

substitute

K =
`(1)/(2) (p^2)/(m)`

from the uncertainty principle,

Δp =
`(h')/(2 \ \Delta x)`

we substitute

K =
`(1)/(2m) ( (h')/(2 \ \Delta x))^2`

`K = (h')/(8 m \ \Delta x^2)`