$A$ microscope using suitable photons is employed to locate an electron in an atom within a distance of $0.1 \, \mathring{A}$. What is the uncertainty involved in the measurement of its velocity?
(N/A) According to Heisenberg's uncertainty principle: $\Delta x \Delta p = \frac{h}{4 \pi}$ or $\Delta x \, m \Delta v = \frac{h}{4 \pi}$.
Rearranging for velocity uncertainty: $\Delta v = \frac{h}{4 \pi \Delta x m}$.
Given values: $h = 6.626 \times 10^{-34} \, J \, s$,$\Delta x = 0.1 \, \mathring{A} = 0.1 \times 10^{-10} \, m$,$m = 9.11 \times 10^{-31} \, kg$.
Substituting the values: $\Delta v = \frac{6.626 \times 10^{-34}}{4 \times 3.1416 \times 0.1 \times 10^{-10} \times 9.11 \times 10^{-31}}$.
Calculation: $\Delta v = 0.579 \times 10^{7} \, m \, s^{-1} = 5.79 \times 10^{6} \, m \, s^{-1}$.
Rearranging for velocity uncertainty: $\Delta v = \frac{h}{4 \pi \Delta x m}$.
Given values: $h = 6.626 \times 10^{-34} \, J \, s$,$\Delta x = 0.1 \, \mathring{A} = 0.1 \times 10^{-10} \, m$,$m = 9.11 \times 10^{-31} \, kg$.
Substituting the values: $\Delta v = \frac{6.626 \times 10^{-34}}{4 \times 3.1416 \times 0.1 \times 10^{-10} \times 9.11 \times 10^{-31}}$.
Calculation: $\Delta v = 0.579 \times 10^{7} \, m \, s^{-1} = 5.79 \times 10^{6} \, m \, s^{-1}$.
Explore More
Similar Questions
The minimum uncertainty in the speed of an electron in a one-dimensional region of length $2 a_{0}$ (where $a_{0} = \text{Bohr radius} = 52.9 \ pm$) is $km \ s^{-1}$. (Given: Mass of electron $= 9.1 \times 10^{-31} \ kg$,Planck's constant $h = 6.63 \times 10^{-34} \ J \ s$)
DifficultJEE MAIN 2022
View SolutionThe uncertainty in the velocities of two particles $A$ and $B$ are $0.03 \ m \ s^{-1}$ and $0.01 \ m \ s^{-1}$ respectively. The mass of $B$ is four times the mass of $A$. The ratio of uncertainties in their positions is
In Heisenberg's uncertainty equation $\Delta x \times \Delta p \ge \frac{h}{4\pi}$,$\Delta p$ stands for:
Easy
View SolutionIf the position of the electron is measured within an accuracy of $\pm 0.002 \,nm$,calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is $\frac{h}{4 \pi \times 0.05 \,nm}$,is there any problem in defining this value?
Difficult
View SolutionWhich of the following statements does not form part of Bohr's model of the hydrogen atom?
MediumAIPMT 1989
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo