Question

A visual inspection of an ant of mass 0.5 mg verifies that it to within an uncertainty of $\mathbf{0}\mathbf{.}\mathbf{7}\mathbf{}\mathit{\mu }\mathit{m}$of a given point, apparently stationary. How fast might the ant actually be moving?

Short answer

The ant moves at a speed of 1.5x10^-22m/s.

Step-by-step solution

Given data.

Mass is $0.5mg\left(\frac{{10}^{-6}kg}{1mg}\right)=5\times {10}^{-7}kg$.

Uncertainty in position $∆x=7\times {10}^{-7}m$.

$\mathcal{h}=1.0545\times {10}^{-34}J·s$.

Heisenberg uncertainty principle.

The following equation describes the Heisenberg uncertainty principle for momentum:

$\mathbf{∆}\mathit{p}\mathbf{∆}\mathit{x}\mathbf{\ge }\frac{\sout{\mathbf{h}}}{\mathbf{2}}$ ……………….. (1)

Momentum.

Knowing that momentum may be stated as p=mv we can derive the following relationship using the equation (1):

$$\begin{gathered} ∆p∆x\ge \frac{\sout{h}}{2} \\ m∆v∆x\ge \frac{\sout{h}}{2} \end{gathered}$$

Uncertainty in velocity.

The formula for velocity uncertainty is then derived:

$$\begin{gathered} ∆p∆x\ge \frac{\sout{h}}{2} \\ ∆v\ge \frac{\sout{h}}{2m∆x} \end{gathered}$$

Velocity of ant.

We can calculate the ant's velocity by solving for$∆v$:

$$\begin{gathered} ∆v\ge \frac{\sout{h}}{2m∆x} \\ ∆v\ge \frac{1.0545\times {10}^{-34}J·s}{2\times \left(5\times {10}^{-7}kg\right)\times \left(7\times {10}^{-7}m\right)} \\ ∆v\ge 1.5\times {10}^{-22}m/s \end{gathered}$$

The ant moves at a speed of $∆v\ge 1.5\times {10}^{-22}m/s$.