An electron is confi ned to a linear region with a length of the same order as the diameter of an atom (about 100 pm). Calculate the minimum uncertainties in its position and speed.

Respuesta :

Explanation:

The minimum uncertainty position is 100 nm.

Therefore, by Heisenberg Uncertainty principal

ΔxΔp≥h/2

Δp≥h/(2Δx)

we know that

h= [tex]1.0546\times10^{-34}[/tex] Js

Δx= [tex]100\times10^{-12} m[/tex]

therefore,

[tex]\Delta p =\frac{1.0546\times10^{-34}}{2\times12\times10^{-12}} = 5.3\times10^{-25} kgms^{-1}[/tex]

therefore,

[tex]\Delta v= \frac{\Delta p}{m} =\frac{5.3\times10^{-25}}{9.11\times10^{-31}}= 5.8\times10^{5} ms^{-1}[/tex]