Conceptual Question on Heisenberg Uncertainty Principle. #2

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 = 10.0 \rm \ pm = 10.0 \times 10^{-12}\rm \ m

To find:

• \Delta P, and
• \Delta V

We know the values:

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

• m_e = 9.11\times 10^{-31}\rm \ kg

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 &\geq \dfrac{h}{4\pi\Delta x}\\ &= \dfrac{6.626 \times 10^{ -34}\rm \ J \cdot s}{4\pi \times (10.0 \times 10^{-12}\rm \ m)}\\&= 5.27 \times 10^{ -24} \rm \  kg \cdot m/s\\\end{aligned}

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

\begin{aligned}\Delta P &= 5.27 \times 10^{ -24} \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_e}\\&= \dfrac{5.27 \times 10^{ -24} \rm \ kg \cdot m/s }{9.11\times 10^{-31}\rm \ kg }\\&= 5.78 \times 10^{6}\rm \ m/s\\\end{aligned}

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

\begin{aligned}\Delta V &= 5.32 \times 10^{6}\rm \ m/s \\\end{aligned}

Given the momentum of the electron is equal to its minimum uncertainty. That is P = \Delta P

So, the kinetic energy of the electron when momentum is equal to its minimum uncertainty is given by

\begin{aligned}K.E. &= \dfrac{P^2}{2m_e}\\&= \dfrac{( \Delta P)^2}{2m_e}\\&= \dfrac{(5.27 \times 10^{ -24} \rm \ kg \cdot m/s)^2}{2 \times (9.11\times 10^{-31}\rm \ kg )}\\&= 1.52 \times 10^{ -17} \rm \ J\\\end{aligned}