Conceptual Question on Heisenberg Uncertainty Principle. #1

Let’s suppose:

• m is the mass of the particle,
• \Delta x is the uncertainty in position, 
• \Delta P is the uncertainty in momentum, and
• \Delta V is the uncertainty in velocity.

Given:

• \Delta x = 2.145 \rm \ nm = 2 \times 10^{-9}\rm \ m
• m = 4.734 \times 10^{ -27}\rm \ kg

To find:

• \Delta P, and
• \Delta V

We know the values:

• h = 6.626 \times 10^{ -34}\rm \  J \cdot s

From the Heisenberg uncertainty principle, the uncertainty in position and the uncertainty in momentum are related by an equation;

\begin{aligned} \Delta x \times \Delta P &\geq \dfrac{h}{4\pi}\\\Rightarrow \Delta P &= \dfrac{h}{4\pi\Delta x}\\ &\geq \dfrac{6.626 \times 10^{ -34}\rm \ : J \cdot s}{4\pi \times 2.145 \times 10^{ -9}\rm \ m}\\&= 2.458 \times 10^{ -27} \rm \  kg \cdot m/s \\\end{aligned}

Therefore, the minimum uncertainty in the particle’s momentum is

\begin{aligned}\Delta P &= 2.458 \times 10^{ -27} \rm \ kg \cdot m/s\\\end{aligned}

The uncertainty in the velocity and the uncertainty in momentum are related as;

\begin{aligned}\Delta P &= m \Delta V\\\end{aligned}

So, the minimum uncertainty in the particle’s velocity is calculated as

\begin{aligned}\Delta V &= \dfrac{\Delta P}{m}\\&= \dfrac{2.458 \times 10^{ -27} \rm \ kg \cdot m/s }{4.734 \times 10^{ -27} \ kg}\\&= 5.620 \times 10^{-1}\rm \ m/s\\\end{aligned}

Therefore, the minimum uncertainty in the particle’s velocity is

\begin{aligned}\Delta V &= 5.620 \times 10^{-1}\rm \ m/s\\\end{aligned}