Question

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)

Answer 1

Answer: $2.89(10)^{-3} m$

Explanation:

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:

$\Delta x \geq \frac{h}{4 \pi m \Delta V}$ (1)

Where:

$\Delta x$ is the uncertainty in the position of the electron

$h=6.626(10)^{-34}J.s$ is the Planck constant

$m=9.11(10)^{-31}kg$ is the mass of the electron

$\Delta V$ is the uncertainty in the velocity of the electron.

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

$\Delta V=2000 m/s(0.001\%)$

$\Delta V=2000 m/s(\frac{0.001}{100})$

$\Delta V=2(10)^{-2} m/s$ (2)

Now, the least possible uncertainty in position $\Delta x_{min}$ is:

$\Delta x_{min}=\frac{h}{4 \pi m \Delta V}$ (3)

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

Finally:

$\Delta x_{min}=2.89(10)^{-3} m$