Conceptual Question on Heisenberg Uncertainty Principle. #4

Let’s suppose:

• m_e is the mass of the electron, 

• \Delta x is the uncertainty in position, 

• \Delta P is the uncertainty in momentum, and 

• \lambda is the electron’s wavelength.

Given:

• \Delta x = 2.00 \times 10^{-8}\rm  \ m

To find:

• \Delta P, and

• \lambda

We know the value:

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

• m_e = 9.11\times 10^{-31}\rm \ kg

Part a.

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}\\ &\geq \dfrac{6.626 \times 10^{ -34}\rm \ : J \cdot s}{4\pi \times (2.00 \times 10^{-8}\rm  \ m)}\\&= 2.64 \times 10^{ -27} \rm \  kg \cdot m/s\\\end{aligned}

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

\begin{aligned}\Delta P &= 2.64 \times 10^{ -27} \rm \  kg \cdot m/s\\\end{aligned}

Part b.

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

Now, from the de -Broglie hypothesis, the electron’s wavelength is calculated as

\begin{aligned}\lambda &= \dfrac{h}{P}\\&= \dfrac{h}{\Delta P}\\&= \dfrac{6.626 \times 10^{ -34}\rm \ : J \cdot s }{2.64 \times 10^{ -27} \rm \  kg \cdot m/s} \\&= 2.51\times 10^{ -7}\rm \ m \\\end{aligned}