Question

An oil drop of mass 10-12 gm is floating on the free surface of a liquid. At any instant the position of the drop can be determined within the error of 10-4 cm. what will be the error in measurement of its velocity.

Answer 1

According to the Heisenberg's uncertainty principle

$$\Delta p\Delta x \ge\dfrac{\hbar}{2}$$

where $\hbar \approx 1.05\times 10^{-34}J\cdot s$ is the reduced Planck constant, $\Delta x = 10^{-4}cm = 10^{-6}m$ is the position error, and $\Delta p$ is the momentum error. Since the mass of the drop is $m = 10^{-12}g = 10^{-15}kg$, then

$$\Delta p = m\Delta v$$

where $\Delta v$ is the error in measurement of drop's velocity. Thus, obtian:

$$m\Delta v\Delta x \ge\dfrac{\hbar}{2} \\ \Delta v \ge\dfrac{\hbar}{2m\Delta x}\\ \Delta v \ge\dfrac{1.05\times 10^{-34}}{2\cdot 10^{-15}\cdot 10^{-6}} = 5.25\times 10^{-14}m/s$$

Answer. $\ge5.25\times 10^{-14}m/s$.