Chapter 7: Problem 114
When an electron is accelerated by a voltage difference, the kinetic energy acquired by the electron equals the voltage times the charge on the electron. Thus, one volt imparts a kinetic energy of \(1.602 \times 10^{-19}\) volt- coulombs, or \(1.602 \times\) \(10^{-19} \mathrm{~J}\). What is the wavelength for electrons accelerated by \(1.00 \times 10^{4}\) volts?
Short Answer
Step by step solution
Understanding Kinetic Energy
Calculate Kinetic Energy
Relate to De Broglie Wavelength
Find Momentum
Calculate Wavelength
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Kinetic Energy
- \( KE \) stands for kinetic energy measured in joules (J).
- \( q \) represents the electron charge, approximately \( 1.602 \times 10^{-19} \) coulombs (C).
- \( V \) is the voltage, the potential difference that accelerates the electron, measured in volts (V).
Momentum
- \( p = \sqrt{2m \cdot KE} \), where:
- \( m \) represents the mass of the electron, approximately \( 9.109 \times 10^{-31} \) kilograms (kg).
- \( KE \) is the kinetic energy of the electron, usually derived from its acceleration by a voltage.
Electron Charge
- Used in the equation \( KE = qV \) to determine the kinetic energy of an electron when accelerated by a voltage.
- Helps in explaining why an electron gains energy when exposed to an electric potential.
- It is a fundamental unit in the study of physics and electronics.
Planck's Constant
- \( \lambda = \frac{h}{p} \), where \( \lambda \) is the wavelength of the particle, \( h \) is Planck's constant, and \( p \) represents momentum.