   Chapter 29, Problem 64PE

Chapter
Section
Textbook Problem

Suppose the velocity of an electron in an atom is known to an accuracy of 2.0 × 10 3 m/s (reasonably accurate compared with orbital velocities). What is the electron's minimum uncertainty in position, and how does this compare with the approximate 0.1-nm size of the atom?

To determine

The minimum uncertainty in position of an electron and the comparison of the value of uncertainty in position with the approximate0.1-nmsize of the atom.

Explanation

Given info:

Uncertainty in velocity is,Δv=2.0×103m/s.

Formula used:

The Heisenberg's Uncertainty principle can be expressed mathematically as,

ΔxΔph4π

Calculation:

The uncertainty in momentum can be calculated as,

Δp=mΔv

The smallest uncertainty in the position of electron will be,

ΔxΔp=h4πΔxmΔv=h4πΔx=h4πmΔv

Substituting the given values in the above equation, we get

Δx=6.626×1034Js4×227×9.11×1031kg×2.0×103m/s=6.626×1034Js229

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