Textbook Question
A parallel-plate capacitor with a 1.0 mm plate separation is charged to 75 V. With what kinetic energy, in eV, must a proton be launched from the negative plate if it is just barely able to reach the positive plate?
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Ch 37: The Foundations of Modern Physics
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796
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Table of contents
- Ch 01: Concepts of Motion35
- Ch 02: Kinematics in One Dimension75
- Ch 03: Vectors and Coordinate Systems58
- Ch 04: Kinematics in Two Dimensions60
- Ch 05: Force and Motion23
- Ch 06: Dynamics I: Motion Along a Line53
- Ch 07: Newton's Third Law27
- Ch 08: Dynamics II: Motion in a Plane52
- Ch 09: Work and Kinetic Energy56
- Ch 10: Interactions and Potential Energy50
- Ch 11: Impulse and Momentum56
- Ch 12: Rotation of a Rigid Body71
- Ch 13: Newton's Theory of Gravity52
- Ch 14: Fluids and Elasticity44
- Ch 15: Oscillations55
- Ch 16: Traveling Waves37
- Ch 17: Superposition39
- Ch 18: A Macroscopic Description of Matter53
- Ch 19: Work, Heat, and the First Law of Thermodynamics60
- Ch 20: The Micro/Macro Connection53
- Ch 21: Heat Engines and Refrigerators34
- Ch 22: Electric Charges and Forces40
- Ch 23: The Electric Field39
- Ch 24: Gauss' Law32
- Ch 25: The Electric Potential63
- Ch 26: Potential and Field57
- Ch 27: Current and Resistance42
- Ch 28: Fundamentals of Circuits39
- Ch 29: The Magnetic Field47
- Ch 30: Electromagnetic Induction60
- Ch 31: Electromagnetic Fields and Waves32
- Ch 32: AC Circuits63
- Ch 33: Wave Optics32
- Ch 34: Ray Optics57
- Ch 35: Optical Instruments30
- Ch 36: Special Relativity38
- Ch 37: The Foundations of Modern Physics25
- Ch 38: Quantization49
- Ch 39: Wave Functions and Uncertainty31
- Ch 40: One-Dimensional Quantum Mechanics35
- Ch 41: Atomic Physics18
- Ch 42: Nuclear Physics27
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 37, Problem 16a
Express in eV (or keV or MeV if more appropriate): The kinetic energy of an electron moving with a speed of 5.0 x 10⁶ m/s .
Verified step by step guidance1
Step 1: Recall the formula for kinetic energy, which is given by \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the electron and \( v \) is its velocity.
Step 2: Substitute the known values into the formula. The mass of an electron \( m \) is approximately \( 9.11 \times 10^{-31} \; \text{kg} \), and the velocity \( v \) is given as \( 5.0 \times 10^6 \; \text{m/s} \).
Step 3: Calculate the kinetic energy in joules using the formula \( KE = \frac{1}{2}mv^2 \). This involves squaring the velocity, multiplying by the mass, and then dividing by 2.
Step 4: Convert the kinetic energy from joules to electronvolts (eV). Use the conversion factor \( 1 \; \text{eV} = 1.602 \times 10^{-19} \; \text{J} \). Divide the kinetic energy in joules by this factor to express it in eV.
Step 5: If the resulting value is very large or very small, express it in keV (\( 1 \; \text{keV} = 10^3 \; \text{eV} \)) or MeV (\( 1 \; \text{MeV} = 10^6 \; \text{eV} \)) for convenience.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion, calculated using the formula KE = 0.5 * m * v², where m is the mass and v is the velocity of the object. For an electron, this energy can be expressed in electronvolts (eV), a unit commonly used in particle physics to represent small energy values.
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Mass of an Electron
The mass of an electron is approximately 9.11 x 10⁻³¹ kg. This small mass is crucial for calculating the kinetic energy of the electron when it is moving at high speeds, as even small changes in velocity can result in significant changes in kinetic energy due to the squared term in the kinetic energy formula.
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Conversion to Electronvolts
Electronvolts (eV) are a unit of energy defined as the amount of kinetic energy gained by an electron when it is accelerated through an electric potential difference of one volt. To convert kinetic energy from joules to electronvolts, one can use the conversion factor 1 eV = 1.602 x 10⁻¹⁹ joules, allowing for easier interpretation of energy values in the context of atomic and subatomic processes.
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Related Practice
Textbook Question
A 0.80-μm-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if the upper electrode is 20 V more positive than the lower electrode. The density of the oil is 885 kg/m3. Does the droplet have a surplus or a deficit of electrons? How many?
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Textbook Question
How many electrons, protons, and neutrons are contained in the following atoms or ions: (a) ¹⁰B, (b) ¹³N⁺, and (c) ¹⁷O⁺⁺⁺?
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Textbook Question
An electron in a cathode-ray beam passes between 2.5-cm-long parallel-plate electrodes that are 5.0 mm apart. A 2.0 mT, 2.5-cm-wide magnetic field is perpendicular to the electric field between the plates. The electron passes through the electrodes without being deflected if the potential difference between the plates is 600 V. If the potential difference between the plates is set to zero, what is the electron's radius of curvature in the magnetic field?
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Textbook Question
Determine the speed of a 15 MeV helium atom.
Textbook Question
Identify the isotope that is 11 times as heavy as ¹²C and has 18 times as many protons as ⁶Li . Give your answer in the form ᴬS, where S is the symbol for the element. See Appendix C: Atomic and Nuclear Data.
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