Question
An electron has a momentum $p=1.7 \times 10^{-25} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s} .$ What is the minimum uncertainty in its position that will keep the relative uncertainty in its momentum $(\Delta p / p)$ below 1.0$\%$ ?
Step 1
Mathematically, this is represented as $\Delta p \Delta x \geq \frac{\hbar}{2\pi}$, where $\Delta p$ is the uncertainty in momentum, $\Delta x$ is the uncertainty in position, and $\hbar$ is the reduced Planck constant. Show more…
Show all steps
Your feedback will help us improve your experience
Zulfiqar Ali and 59 other Physics 103 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
The uncertainty in an electron's position is $0.15 \mathrm{nm}$. (a) What is the minimum uncertainty $\Delta p$ in its momentum? (b) What is the kinetic energy of an electron whose momentum is equal to this uncertainty $(\Delta p=p) ?$
The uncertainty an electron's position is 0.010 nm.(a) What is the minimum uncertainty $\Delta p$ in its momentum? (b) What is the kinetic energy of an electron whose momentum is equal to this uncertainty $(\Delta p=p) ?$
Suppose that an electron is trapped within a small region and the uncertainty in its position is $3.0 \times 10^{-15} \mathrm{~m}$. What is the minimum uncertainty in the electron's momentum?
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD