Question

If the electron's position is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the electron's momentum. Suppose the momentum of the electron is \(\frac{h}{4\pi m}\) × 0.05 nm, is there any problem in defining this value?

Answer 1

From Heisenberg's uncertainty principle,
Δx × Δp =\(\frac{h}{4\pi}\) ⇒ Δp = \(\frac{1}{\Delta x}\) . \(\frac{h}{4\pi}\)
Where, Δx = uncertainty in the position of the electron
Δp = uncertainty in momentum of the electron
Substituting the values in the expression of Δp:
Δp = \(\frac{1}{0.002}\)nm × \(\frac{6.626 × 10^{-34 Js}}{ 4 × (3.14)}\)
= 2.637 × 10^-23 Js^-1
Δp = 2.637 × 10²³ kgms (1 J = 1 kgms 2s)
∴ Uncertainty in the momentum of the electron = 2.637 × 10²³ kgms^-1.
Actual momentum = \(\frac{h}{4\pi m}\)× 0.05 nm
=\(\frac{ 6.626 × 10^{-34} Js}{4 × 3.14 × 5.0 × 10^ {-11} m}\)
= 1.055 × 10^-24 kgms^-1
Since the magnitude of the actual momentum is smaller than the uncertainty, the value cannot be defined .