Textbook Question
What is the wavelength, in nm, of a photon with energy (a) 0.30 eV, (b) 3.0 eV, and (c) 30 eV? For each, is this wavelength visible, ultraviolet, or infrared light?
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Ch 38: Quantization
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 38, Problem 13
A 100 W incandescent lightbulb emits about 5 W of visible light. (The other 95 W are emitted as infrared radiation or lost as heat to the surroundings.) The average wavelength of the visible light is about 600 nm, so make the simplifying assumption that all the light has this wavelength. How many visible-light photons does the bulb emit per second?
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Step 1: Start by calculating the energy of a single photon using the formula \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J·s} \)), \( c \) is the speed of light (\( 3.00 \times 10^8 \, \text{m/s} \)), and \( \lambda \) is the wavelength of the light (600 nm or \( 600 \times 10^{-9} \, \text{m} \)).
Step 2: Substitute the given values into the formula \( E = \frac{hc}{\lambda} \) to calculate the energy of a single photon. Ensure that the wavelength is converted to meters before substituting.
Step 3: Determine the total energy emitted as visible light per second. This is given as 5 W, which means 5 joules of energy are emitted per second.
Step 4: Calculate the number of photons emitted per second by dividing the total energy emitted as visible light per second by the energy of a single photon. Use the formula \( N = \frac{P}{E} \), where \( P \) is the power of visible light (5 W) and \( E \) is the energy of a single photon.
Step 5: Simplify the expression to find the number of photons emitted per second. Ensure that the units are consistent throughout the calculation (e.g., joules for energy).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Photon Energy
The energy of a photon is determined by its wavelength, given by the equation E = hc/λ, where E is energy, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength in meters. For visible light at 600 nm, this relationship allows us to calculate the energy of each photon emitted by the lightbulb.
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Power and Energy Relationship
Power is defined as the rate at which energy is transferred or converted, measured in watts (W). In this context, the lightbulb emits 5 W of visible light, meaning it produces 5 joules of energy per second. This relationship is crucial for determining how many photons are emitted based on the energy each photon carries.
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Calculating Photon Emission Rate
To find the number of photons emitted per second, divide the total power output in visible light (5 W) by the energy of a single photon. This calculation provides the photon emission rate, which is essential for understanding the light output of the bulb in terms of individual photons.
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Related Practice
Textbook Question
Photoelectrons are observed when a metal is illuminated by light with a wavelength less than 388 nm. What is the metal's work function?
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Textbook Question
At what speed is an electron’s de Broglie wavelength (a) 1.0 nm, (b) 1.0 μm, and (c) 1.0 mm?
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Textbook Question
A photoelectric-effect experiment finds a stopping potential of 1.56 V when light of 200 nm is used to illuminate the cathode. From what metal is the cathode made?
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Textbook Question
What is the energy, in keV, of 75 keV x-ray photons that are backscattered (i.e., scattered directly back toward the source) by the electrons in a target?
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Textbook Question
55 keV x-ray photons are incident on a target. At what scattering angle do the scattered photons have an energy of 50 keV?
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