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 .