Chapter 39: Problem 20
A hydrogen atom initially in its ground level absorbs a photon, which excites the atom to the \(n\) = 3 level. Determine the wavelength and frequency of the photon.
Short Answer
Expert verified
The photon's wavelength is 102.8 nm, and its frequency is \(2.918 \times 10^{15}\) Hz.
Step by step solution
01
Understanding the Energy Levels of Hydrogen Atom
The energy levels of a hydrogen atom are given by the formula \( E_n = -13.6 \text{ eV} \times \frac{1}{n^2} \), where \( n \) is the principal quantum number. For the ground state, \( n = 1 \), and for the excited state, \( n = 3 \).
02
Calculate Energy in Ground State (n=1)
Substitute \( n = 1 \) into the energy level equation: \[ E_1 = -13.6 \text{ eV} \times \frac{1}{1^2} = -13.6 \text{ eV}. \]
03
Calculate Energy in Excited State (n=3)
Substitute \( n = 3 \) into the energy level equation: \[ E_3 = -13.6 \text{ eV} \times \frac{1}{3^2} = -13.6 \text{ eV} \times \frac{1}{9} = -1.51 \text{ eV}. \]
04
Determine the Energy Change
The energy absorbed by the photon is the difference between the energies of the two states. \[ \Delta E = E_3 - E_1 = -1.51 \text{ eV} - (-13.6 \text{ eV}) = 12.09 \text{ eV}. \]
05
Convert Energy from eV to Joules
Since energy in photon calculations is often used in joules, convert the energy from eV to joules. \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \), so: \[ \Delta E = 12.09 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV} = 1.935 \times 10^{-18} \text{ J}. \]
06
Calculate the Frequency of the Photon
Use the formula \( E = h u \), where \( h = 6.626 \times 10^{-34} \text{ Js} \) is Planck's constant. Solve for \( u \): \[ u = \frac{\Delta E}{h} = \frac{1.935 \times 10^{-18} \text{ J}}{6.626 \times 10^{-34} \text{ Js}} = 2.918 \times 10^{15} \text{ Hz}. \]
07
Calculate the Wavelength of the Photon
Use the wave equation \( \lambda = \frac{c}{u} \), where \( c = 3 \times 10^8 \text{ m/s} \) is the speed of light. \[ \lambda = \frac{3 \times 10^8 \text{ m/s}}{2.918 \times 10^{15} \text{ Hz}} = 1.028 \times 10^{-7} \text{ m} = 102.8 \text{ nm}. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Photon Absorption
When a hydrogen atom absorbs a photon, it gains energy and can become excited. This means an electron within the atom jumps to a higher energy level. In our scenario, the atom initially in its ground state (
=1
equal 1) absorbs energy and moves to the
=3
equal 3 level. This transition occurs because the photon's energy precisely matches the energy difference between these two levels.
Photon absorption is crucial in understanding how light interacts with matter. It explains phenomena such as fluorescence, where absorbed light is re-emitted at different wavelengths.
Understanding photon absorption helps explain the behaviors observed in many physical systems, such as how sunlight powers photovoltaic cells or how plants use light for photosynthesis.
Photon absorption is crucial in understanding how light interacts with matter. It explains phenomena such as fluorescence, where absorbed light is re-emitted at different wavelengths.
Understanding photon absorption helps explain the behaviors observed in many physical systems, such as how sunlight powers photovoltaic cells or how plants use light for photosynthesis.
Wavelength and Frequency Calculation
Determining the characteristics of the absorbed photon involves calculating both its wavelength and frequency. When energy is calculated, it can be further broken down based on these properties.
To find a photon's frequency, which is how many waves pass a point per second, we use the relation between energy and frequency: E = h \cdot fE=hf, where \( E \) is the energy, \( h \) is Planck's constant, and \( f \) is frequency. Solving for frequency, we get \( f = \frac{E}{h} \).
Wavelength calculation involves using the speed of light \( c = 3 \times 10^8\text{ m/s} \) as the speed at which electromagnetic waves travel. The wavelength is derived from the relationship \( \lambda = \frac{c}{f} \), linking wavelength \( \lambda \), speed of light \( c \), and frequency \( f \). Thus, finding these characteristics offers insights into the nature of light that has been absorbed or emitted.
To find a photon's frequency, which is how many waves pass a point per second, we use the relation between energy and frequency: E = h \cdot fE=hf, where \( E \) is the energy, \( h \) is Planck's constant, and \( f \) is frequency. Solving for frequency, we get \( f = \frac{E}{h} \).
Wavelength calculation involves using the speed of light \( c = 3 \times 10^8\text{ m/s} \) as the speed at which electromagnetic waves travel. The wavelength is derived from the relationship \( \lambda = \frac{c}{f} \), linking wavelength \( \lambda \), speed of light \( c \), and frequency \( f \). Thus, finding these characteristics offers insights into the nature of light that has been absorbed or emitted.
Quantum Transitions in Hydrogen
In hydrogen's simple atom, electrons transition between defined energy levels or quantum states. We call these movements quantum transitions. They occur when specific amounts of energy, like a photon's energy, add or take away energy from the electron.
The Schroedinger equation predicts possible electron states, giving rise to discrete energy levels. For example, moving from \( n=1 \) to \( n=3 \) requires a specific energy, calculable via changes in electron energy levels: \( E_n = -\frac{13.6\text{ eV}}{n^2} \). The precise energy change between levels (known as quantum transition) guides what wavelength or frequency photons are emitted or absorbed.
Quantum transitions also explain atomic spectra, showing light as lines at precise energies, unique for each element. This characteristic remains a cornerstone in identifying elements via their spectral lines in stars and gas clouds.
The Schroedinger equation predicts possible electron states, giving rise to discrete energy levels. For example, moving from \( n=1 \) to \( n=3 \) requires a specific energy, calculable via changes in electron energy levels: \( E_n = -\frac{13.6\text{ eV}}{n^2} \). The precise energy change between levels (known as quantum transition) guides what wavelength or frequency photons are emitted or absorbed.
Quantum transitions also explain atomic spectra, showing light as lines at precise energies, unique for each element. This characteristic remains a cornerstone in identifying elements via their spectral lines in stars and gas clouds.
Planck's Constant
Planck’s constant is a fundamental quantity in quantum mechanics, symbolized by \( h \), approximately equal to \( 6.626 \times 10^{-34} \text{ Js} \). It relates the energy of photons to their frequency, providing a key to unlocking quantum behaviors.
The relationship \( E = hf \) allows us to calculate how energy and frequency interconnect for photons. This principle underlies numerous technologies, such as modern electronics and lasers. Planck's introduction of this constant signaled the start of quantum physics, changing how scientists perceive the micro-universe.
Planck's constant remains vital in modern science, appearing in formulas calculating the exchange of energy on very small scales.
The relationship \( E = hf \) allows us to calculate how energy and frequency interconnect for photons. This principle underlies numerous technologies, such as modern electronics and lasers. Planck's introduction of this constant signaled the start of quantum physics, changing how scientists perceive the micro-universe.
Planck's constant remains vital in modern science, appearing in formulas calculating the exchange of energy on very small scales.
Energy Conversion
Energy conversion is the process of changing energy from one form to another. In our hydrogen atom example, energy shifts forms during an electron's transition between energy levels.
To express energy accurately, it often requires conversion from one unit to another, such as from electronvolts (eV) to joules (J), using the conversion factor \( 1\text{ eV} = 1.602 \times 10^{-19}\text{ J} \). This step is crucial to apply the energy in different contexts, like calculating frequency in joules per second (Js).
Awareness of energy conversion helps grasp how different types of energies interact in varied systems, from atomic scales in physics to larger systems in engineering, where energy forms transformation underpins electricity generation or solar power utilization.
To express energy accurately, it often requires conversion from one unit to another, such as from electronvolts (eV) to joules (J), using the conversion factor \( 1\text{ eV} = 1.602 \times 10^{-19}\text{ J} \). This step is crucial to apply the energy in different contexts, like calculating frequency in joules per second (Js).
Awareness of energy conversion helps grasp how different types of energies interact in varied systems, from atomic scales in physics to larger systems in engineering, where energy forms transformation underpins electricity generation or solar power utilization.