What is the value of $\Delta v \cdot \Delta x$ for an electron? What does it indicate?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) According to Heisenberg's uncertainty principle,$\Delta x \cdot \Delta p \ge \frac{h}{4\pi m}$.
Substituting $\Delta p = m \cdot \Delta v$,we get $\Delta x \cdot \Delta v \ge \frac{h}{4\pi m}$.
For an electron,$m = 9.11 \times 10^{-31} \ kg$ and $h = 6.626 \times 10^{-34} \ J \cdot s$.
$\Delta x \cdot \Delta v \ge \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.11 \times 10^{-31}} \approx 5.79 \times 10^{-5} \ m^2 \ s^{-1}$.
This value indicates that for an electron,it is impossible to determine both the position and velocity simultaneously with absolute precision.

Explore More

Similar Questions

What is indicated by $\psi$ for atomic and molecular orbitals? How is the molecular orbital obtained by the Schrodinger wave equation?

Which of the following statements does not form part of Bohr's model of the hydrogen atom?

For an electron,if the uncertainty in velocity is $\Delta \nu$,the uncertainty in its position $(\Delta x)$ is given by:

Heisenberg's uncertainty principle is not applicable to which of the following?

Which one is not the correct relation in the following?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo