Domanda del libro di testo
One type of sunburn occurs on exposure to UV light of wavelength in the vicinity of 325 nm. (a) What is the energy of a photon of this wavelength?
Ch.6 - Electronic Structure of Atoms
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232
Non è quello che usi tu?Cambia libro di testoScegli un libro di testo diverso
TOC.tableOfContents
- Ch.1 - Introduction: Matter, Energy, and Measurement177
- Ch.2 - Atoms, Molecules, and Ions231
- Ch.3 - Chemical Reactions and Reaction Stoichiometry216
- Ch.4 - Reactions in Aqueous Solution189
- Ch.5 - Thermochemistry175
- Ch.6 - Electronic Structure of Atoms168
- Ch.7 - Periodic Properties of the Elements150
- Ch.8 - Basic Concepts of Chemical Bonding177
- Ch.9 - Molecular Geometry and Bonding Theories200
- Ch.10 - Gases179
- Ch.11 - Liquids and Intermolecular Forces113
- Ch.12 - Solids and Modern Materials154
- Ch.13 - Properties of Solutions157
- Ch.14 - Chemical Kinetics199
- Ch.15 - Chemical Equilibrium128
- Ch.16 - Acid-Base Equilibria164
- Ch.17 - Additional Aspects of Aqueous Equilibria163
- Ch.18 - Chemistry of the Environment105
- Ch.19 - Chemical Thermodynamics164
- Ch.20 - Electrochemistry171
- Ch.21 - Nuclear Chemistry122
- Ch.22 - Chemistry of the Nonmetals82
- Ch.23 - Transition Metals and Coordination Chemistry79
- Ch.24 - The Chemistry of Life: Organic and Biological Chemistry16
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232Non è quello che usi tu?Cambia libro di testo
Capitolo 6, Problema 29b
One type of sunburn occurs on exposure to UV light of wavelength in the vicinity of 325 nm. (b) What is the energy of a mole of these photons?
Guida verificata passo dopo passo1
First, understand that the energy of a photon can be calculated using the equation: \( E = \frac{hc}{\lambda} \), where \( E \) is the energy of the photon, \( h \) 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.
Convert the given wavelength from nanometers to meters. Since 1 nm = \( 10^{-9} \) m, the wavelength \( \lambda = 325 \text{ nm} \) is equivalent to \( 325 \times 10^{-9} \text{ m} \).
Substitute the values of \( h \), \( c \), and \( \lambda \) into the energy equation: \( E = \frac{(6.626 \times 10^{-34} \text{ J s})(3.00 \times 10^8 \text{ m/s})}{325 \times 10^{-9} \text{ m}} \).
Calculate the energy of a single photon using the equation from the previous step. This will give you the energy in joules for one photon.
To find the energy of a mole of photons, multiply the energy of a single photon by Avogadro's number \( (6.022 \times 10^{23} \text{ mol}^{-1}) \). This will give you the energy in joules per mole of photons.
Risposta video verificata per un problema simile:
Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
7mConcetti chiave
Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.
Photon Energy
Photon energy is the energy carried by a single photon, which can be calculated using 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 (3.00 x 10^8 m/s), and λ is the wavelength in meters. For UV light at 325 nm, this formula allows us to determine the energy of individual photons.
Video consigliato:
Percorso guidato
01:40
Photon Energy Formulas
Mole Concept
The mole is a fundamental unit in chemistry that quantifies the amount of substance. One mole contains Avogadro's number of entities, approximately 6.022 x 10^23. When calculating the energy of a mole of photons, we multiply the energy of a single photon by Avogadro's number to find the total energy associated with that quantity of photons.
Video consigliato:
Percorso guidato
01:25Mole Concept
Wavelength and Energy Relationship
The relationship between wavelength and energy is inversely proportional; as the wavelength decreases, the energy of the photon increases. This principle is crucial in understanding electromagnetic radiation, including UV light. Shorter wavelengths, like 325 nm, correspond to higher energy photons, which can have significant biological effects, such as causing sunburn.
Video consigliato:
Percorso guidato
00:31Frequency-Wavelength Relationship
Pratica correlata
Domanda del libro di testo
One type of sunburn occurs on exposure to UV light of wavelength in the vicinity of 325 nm. (d) These UV photons can break chemical bonds in your skin to cause sunburn—a form of radiation damage. If the 325-nm radiation provides exactly the energy to break an average chemical bond in the skin, estimate the average energy of these bonds in kJ/mol.
1890
views
Domanda del libro di testo
An AM radio station broadcasts at 1000 kHz and its FM partner broadcasts at 100 MHz. Calculate and compare the energy of the photons emitted by these two radio stations.
905
views
Domanda del libro di testo
One type of sunburn occurs on exposure to UV light of wavelength in the vicinity of 325 nm. (c) How many photons are in a 1.00 mJ burst of this radiation?
926
views
Domanda del libro di testo
The energy from radiation can be used to rupture chemical bonds. A minimum energy of 192 kJ/mol is required to break the bromine–bromine bond in Br2. What is the longest wavelength of radiation that possesses the necessary energy to break the bond? What type of electromagnetic radiation is this?
1758
views
2
rank