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 sp…

Question

asked Aug 10, 2021 212k views

1 Answer

Danyim · Answer 1 · 2021-08-14T22:23:40+0000

Step-by-step explanation:

The minimum uncertainty position is 100 nm.

Therefore, by Heisenberg Uncertainty principal

ΔxΔp≥h/2

Δp≥h/(2Δx)

we know that

h=
1.0546*10^(-34) Js

Δx=
100*10^(-12) m

therefore,

$\Delta p =(1.0546*10^(-34))/(2*12*10^(-12)) = 5.3*10^(-25) kgms^(-1)$

therefore,

$\Delta v= (\Delta p)/(m) =(5.3*10^(-25))/(9.11*10^(-31))= 5.8*10^(5) ms^(-1)$