   Chapter 29, Problem 62PE

Chapter
Section
Textbook Problem

(a) If the position of an electron in a membrane is measured to an accuracy of 1.00 μm , what is the electron's minimum uncertainty in velocity? (b) If the electron has this velocity, what is its kinetic energy in eV? (c) What are the implications of this energy, comparing it to typical molecular binding energies?

To determine

(a)

The minimum uncertainty in the velocity of electron.

Explanation

Given info:

Uncertainty in the position of electron is,Δx=1.0μm.

Mass of the electron is,m=9.11×1031kg.

Formula used:

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

ΔxΔph4π

Calculation:

Substituting the given values in the above equation, we get

Δp=h4π×ΔxΔp=6.626×1034Js4×3.14×1.0×106mΔp=5

To determine

(b)

The kinetic energy of electron ineV.

To determine

(c)

The implications of the kinetic energy comparing to typical molecular binding energies.

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