Question

calculate the uncertainty in velocity of an electron at a distance of 0.01 nm from the nucleus?​

Answer 1

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