Suppose that the speed of an electron traveling 2.0 km/s is known to an accuracy of 1 part in 105 (i.e., within 0.0010%). What is the least possible uncertainty within which we can determine the position of this electron? (melectron = 9.11 × 10-31 kg, h = 6.626 × 10-34 J ∙ s)

The Heisenberg uncertainty principle postulates that the fact each particle has a wave associated with it, imposes restrictions on the ability to determine its position and speed at the same time.

In other words:

It is impossible to measure simultaneously (according to quantum physics), and with absolute precision, the value of the position and the momentum (linear momentum) of a particle. Thus, in general, the greater the precision in the measurement of one of these magnitudes, the greater the uncertainty in the measure of the other complementary variable.

Mathematically this principle is written as:

[tex]\Delta x \geq \frac{h}{4 \pi m \Delta V}[/tex] (1)

Where:

[tex]\Delta x[/tex] is the uncertainty in the position of the electron

[tex]h=6.626(10)^{-34}J.s[/tex] is the Planck constant

[tex]m=9.11(10)^{-31}kg[/tex] is the mass of the electron

[tex]\Delta V[/tex] is the uncertainty in the velocity of the electron.

If we know the accuracy of the velocity is [tex]0.001\%[/tex] of the velocity of the electron [tex]V=2 km/s=2000 m/s[/tex], then [tex]\Delta V[/tex] is:

[tex]\Delta V=2000 m/s(0.001\%)[/tex]

[tex]\Delta V=2000 m/s(\frac{0.001}{100})[/tex]

[tex]\Delta V=2(10)^{-2} m/s[/tex] (2)

Now, the least possible uncertainty in position [tex]\Delta x_{min}[/tex] is:

[tex]\Delta x_{min}=\frac{h}{4 \pi m \Delta V}[/tex] (3)

[tex]\Delta x_{min}=\frac{6.626(10)^{-34}J.s}{4 \pi (9.11(10)^{-31}kg) (2(10)^{-2} m/s)}[/tex] (4)

Finally:

[tex]\Delta x_{min}=2.89(10)^{-3} m[/tex]