Question

If an electron is to be located with in 5 x 10-5 A0

, what will be the uncertainity in its

velocity?​

Answer 1

Δx = 5 x 10^-15 m

Δx.Δv = h / 4π.m                 (where h= 6.626 x 10^-34 and m= 9.109×10^-31 )

∴  Δv = h / 4π.m.Δx

Δv = 6.626 x 10^-34 / 4 x 3.14 x 9.109 x 10^-31 x 5 x 10^-5

Δv = 1.158 x 10^10 : Ans (approx)

There will be an uncertainty of 1.158 x 10^10 (approx) in its velocity.

Hope I Helped

Answer 2

Explanation:

◇v= ___h_____

4pie ×m×◇x

given =◇x= 5×10^-5A

in change into m =5×10^-15 m

, then put the value

◇v = 5.25×10^-35

____________

9.1×10^-31×5×10^-15 m ( h /4pie =5.25×10^-35)

( m = 9.1×10^-31)

◇v = 1.15×10^10 .

thus , delta v = 1.15×10^10 .