2.61 If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h/4πm × 0.05 nm, is there any problem in defining this value.


From Heisenberg’s uncertainty principle, 

x × p = h4π  p = 1x·h4π

Where,
∆x = uncertainty in position of the electron
∆p = uncertainty in momentum of the electron
Substituting the values in the expression of ∆p:
p = 10.002 nm x 6.626 x 10-34 Js4 x 3.14
= 12 x 10-12 m x 6.626 x 10-34 Js4 x 3.14
= 2.637 × 10–23 Jsm–1
∆p = 2.637 × 10–23 kgms–1
(1 J = 1 kgms2s–1)
Uncertainty in the momentum of the electron = 2.637 × 10–23 kgms–1.

Actual momentum = h4πm x 0.05 nm

6.626 x 10-34 Js4 x 3.14 x 5.0 x 10-11 m

= 1.055 × 10–24 kgm/s

Since the magnitude of the actual momentum is smaller than the uncertainty, the value cannot be defined.