Question

if the uncertainity of the velocity of an electron is 0.1%of the velocity of light. calculate the uncertainity in the position of the electron​

Answer 1

Answer:

According to Heisenberg's uncertainity principle:

Δx.Δp=

4π

h

Δx.Δ(mv)=

4π

h

Δv=

4πmΔx

h

h=6.626×10

−34

Js=Planck's constant

Given m=mass of electron=9.1×10

−31

kg and Δx=1000

A

˚

=10

−7

m

upon substitution we get:

Δv=

4×3.14×9.1×10

−31

×10

−7

6.626×10

−34

Δv=0.0579×10

4

=5.79×10

2

m/s