**Educational Content on Uncertainty in Quantum Mechanics****Problem Statement:**What is the minimum uncertainty in position, in millimeters, of an electron whose velocity is known to be between 2.96×10^5 m/s and 3.08×10^5 m/s?**Task:**Express your answer using two significant figures.**User Input Interface:**- There is a text box where users can enter their answer. It includes formatting options such as square roots, fractions, and scientific notation for clarity and precision.- The units for the answer input are in millimeters (mm).**Current User Input:**- The user has entered a value of "9.7" mm as their answer.**Feedback Mechanism:**- Upon submission, the user received feedback indicating that the answer was incorrect.- The system provides a prompt: "Incorrect; Try Again; 5 attempts remaining."**Instructional Guidance:**To find the minimum uncertainty in position, one can use the Heisenberg Uncertainty Principle, which relates the uncertainties in position (\(\Delta x\)) and velocity (\(\Delta v\)) of a particle:\[ \Delta x \cdot m \cdot \Delta v \geq \frac{h}{4\pi} \]where:- \( \Delta x \) is the uncertainty in position.- \( m \) is the mass of the electron (\(9.11 \times 10^{-31}\) kg).- \( \Delta v \) is the uncertainty in velocity (\(3.08 \times 10^5 - 2.96 \times 10^5\) m/s).- \( h \) is Planck's constant (\(6.626 \times 10^{-34}\) Js).Use this equation to calculate the minimum uncertainty in position, ensuring the result is in millimeters and expresses the answer to two significant figures.

Given, V1 = 2.96 x 105 m/s V2 = 3.08 x 105 m/s (i) Uncertainty in velocity (ΔV) = (3.08 -2.96)x 105 ΔV = 0.12 x 105 m/s (ii) Uncertainty in momentum (Δp) = m ΔV m = 9.1 x 10-31 kg Δp = (9.1 x 0.12) x 10-26 = 1.09 x 10-26 kg-m/s (iii) According to Heisenberg principle, Δx . Δp = h/2π Δx = h/(2π . Δp) = (6.6 x 10-34)/(2 x 3.14 x 1.09 x 10-26) = 0.96 x 10-9 Δx = 0.96 nm (minimum uncertainty in posi)

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What is the minimum uncertainty in position, in nm, of an electron whose velocity is known to be between 2.96x105 m/s and 3.08×105 m/s ? Express your answer using two significant figures. Vη ΑΣφ x" х.10п 9.7 nm Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining

What is the minimum uncertainty in position, in nm, of an electron whose velocity is known to be between 2.96x105 m/s and 3.08×105 m/s ? Express your answer using two significant figures. Vη ΑΣφ x" х.10п 9.7 nm Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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