Ncert Exercises & Solutions

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,

$∆ x \times ∆ p = \frac{h}{4 π} \Rightarrow ∆ p = \frac{1}{∆ x} \cdot \frac{h}{4 π}$

Where,
∆x = uncertainty in position of the electron
∆p = uncertainty in momentum of the electron
Substituting the values in the expression of ∆p:
$∆ p = \frac{1}{0.002 nm} x \frac{6.626 x 10^{- 34 Js}}{4 x (3.14)}$
$= \frac{1}{2 x 10^{- 12} m} x \frac{6.626 x 10^{- 34} Js}{4 x (3.14)}$
= 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 = $\frac{h}{4 π_{m} x 0.05 nm}$

= $\frac{6.626 x 10^{- 34} Js}{4 x 3.14 x 5.0 x 10^{- 11} m}$

= 1.055 × 10–24 kgm/s

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

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