Matter: hot or cold
| Module 5: Rise and the fall of the clockwork universe 5.2 Matter |
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| 5.2.2 |
Matter: hot or cold I) Ratios of numbers of particles in quantum states of different energy, at different temperatures (classical approximation only) II) Qualitative effects of temperature in processes with an activation energy (for example, changes of state, thermionic emission, ionization, conduction in semiconductors, viscous flow). b) Make appropriate use of: by sketching and interpreting: I) Graphs showing the variation of the Boltzmann factor with energy and temperature. c) Make calculations and estimates involving: I) Ratios of characteristic energies to the energy [math]kT[/math] II) Boltzmann factor |
a) Describe and explain:
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I) Ratios of Numbers of Particles in Quantum States of Different Energy (Classical Approximation)
- ⇒ Background:
- In a system with particles distributed among quantum states of different energies, the ratio of particles in two states depends on the temperature and the energy difference between the states.
- This distribution is given by the Boltzmann factor in the classical approximation.
- ⇒ Formula:
- [math]N_2/N_1 = e^{-\frac{\Delta E}{kT}}[/math]
- Where:
- - [math]N_2[/math]: Number of particles in the higher energy state.
- – [math]N_1[/math]: Number of particles in the lower energy state.
- – ΔE: Energy difference between the two states (J).
- – [math]k = 1.38 × 10^{-23} J/K[/math]: Boltzmann constant
- – T: Absolute temperature (K).
- ⇒ Explanation:
- 1. Temperature and Population Ratios:
- At low temperatures (T is small), [math]\frac{\Delta E}{kT}[/math] is large, so [math]e^{-\frac{\Delta E}{kT}}[/math] is small. This means most particles are in the lower energy state ([math]N_1 ≫ N_2[/math]).
- At high temperatures (T is large), becomes smaller, [math]\frac{\Delta E}{kT}[/math] so [math]e^{-\frac{\Delta E}{kT}}[/math] approaches 1. This means particles are more evenly distributed between the two states.
- 2. Physical Interpretation:
- The ratio [math]\frac{N_2}{N_1}[/math] represents the likelihood of finding particles in the higher energy state compared to the lower energy state.
- Higher temperatures provide more energy, allowing more particles to occupy higher energy states.
- ⇒ Example:
- Energy difference:[math]∆E = 2.5 × 10^{-20} J[/math].
- Temperature: T=300 K.
- [math]\frac{N_2}{N_1} = e^{-\frac{\Delta E}{kT}} \\ \frac{N_2}{N_1} = e^{-\frac{(2.5 \times 10^{-20})}{(1.38 \times 10^{-23}) (300)}} \\ \frac{N_2}{N_1} = e^{-0.604} \\ \frac{N_2}{N_1} \approx 0.546[/math]
- Interpretation: At 300 K, approximately 54.6% as many particles are in the higher energy state as in the lower energy state.
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II) Qualitative Effects of Temperature in Processes with Activation Energy
- ⇒ Activation Energy:
- Activation energy ([math]E_a[/math]) is the minimum energy required for a process to occur.
- Examples include:
- – Phase changes (e.g., melting, boiling).
- – Thermionic emission (release of electrons from a hot surface).
- – Ionization (removal of electrons from atoms).
- – Semiconductor conduction (exciting electrons to the conduction band).
- – Viscous flow (movement in liquids).
- Figure 1 Activation energy
- ⇒ Principle:
- At higher temperatures, particles have more kinetic energy. The fraction of particles with energy equal to or greater than the activation energy increases, making processes with activation energy more likely.
- ⇒ Qualitative Effects of Temperature:
- 1. Phase Changes:
- – Melting and boiling involve overcoming intermolecular forces.
- – At higher temperatures, more particles have sufficient energy to overcome these forces, resulting in phase transitions.
- Figure 2 Phase change
- 2. Thermionic Emission:
- – Electrons are bound to the surface of metals with a certain energy barrier ([math]E_a[/math]).
- – At higher temperatures, more electrons acquire enough energy to escape, increasing the emission rate.
- 3. Ionization:
- – To ionize an atom, electrons must gain sufficient energy to escape their orbits.
- Higher temperatures increase the fraction of electrons with enough energy to ionize atoms.
- 4. Conduction in Semiconductors:
- – Semiconductors require electrons to move from the valence band to the conduction band ([math]\text{gap energy} E_g[/math]).
- – Higher temperatures increase the number of electrons with sufficient energy to cross this gap, enhancing conductivity.
- 5. Viscous Flow:
- – In liquids, molecules must overcome intermolecular forces to move relative to one another.
- – As temperature increases, the fraction of molecules with sufficient energy to flow increases, reducing viscosity.
- ⇒ Mathematical Context:
- For processes dependent on activation energy, the rate (R) is often modeled as:
- [math]R \propto e^{-\frac{E_a}{kT}}[/math]
- – Higher T: Exponent becomes less negative, and R
- – Lower T: Exponent becomes more negative, and R
- ⇒ Examples:
- 6. Thermionic Emission:
- A metal surface at 500 K may emit very few electrons.
- Increasing the temperature to 1000 K significantly boosts electron emission due to the exponential dependence on T.
- 7. Semiconductors:
- At low temperatures, semiconductors behave as insulators (few electrons in the conduction band).
- At higher temperatures, conductivity increases because more electrons gain enough energy to jump the band gap.
b) Make appropriate use of: by sketching and interpreting:
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I) Use of Graphs: Variation of the Boltzmann Factor with Energy and Temperature
- ⇒ The Boltzmann Factor Overview:
- The Boltzmann factor describes the relative probability of particles occupying different energy states in a system at thermal equilibrium. It is given by:
- [math]f(E) = e^{-\frac{E}{kT}}[/math]
- Where:
- – [math]f(E) [/math]: Probability of a particle being in a state with energy
- – E: Energy of the state (J).
- – [math]k = 1.38 × 10^{-23} J/K[/math]: Boltzmann constant.
- T: Absolute temperature (K).
- ⇒ How the Boltzmann Factor Changes with Energy and Temperature
- 1. Dependence on Energy (E):
- At a fixed temperature, the Boltzmann factor decreases exponentially as E
- Higher energy states are less likely to be occupied because fewer particles have sufficient energy.
- 2. Dependence on Temperature (T):
- At higher temperatures, the Boltzmann factor decreases less steeply with E, indicating a higher probability of particles occupying higher energy states.
- At lower temperatures, the Boltzmann factor decreases very steeply with E, meaning particles are more likely to remain in lower energy states.
- ⇒ Graphical Representation
- 1. Graph Showing Variation of [math]f(E)[/math] with Energy (E) at Different Temperatures (T)
- Horizontal axis: Energy (E).
- Vertical axis: Boltzmann factor [math]f(E) = e^{-\frac{E}{kT}}[/math]
- ⇒ Observations:
- 1. At Low Temperature:
- For a small T, the factor [math]e^{-\frac{E}{kT}}[/math]falls sharply with increasing E.
- Most particles are concentrated in low-energy states.
- 2. At High Temperature:
- For a large T, the factor [math]e^{-\frac{E}{kT}}[/math] decreases more gradually.
- A significant fraction of particles occupy higher-energy states.
- 3. Interpretation of the Graph
Temperature |
Graph behavior |
Physical Meaning |
Low T |
Rapid drop on f(E) with E |
Few particles have sufficient energy to reach high-energy states |
High T |
Gradual drop in f(E) with E |
More particles have enough energy to populate high-energy states. |
- ⇒ Graph Example:
- For visualization:
- – At [math]T_1 = 300 K[/math], f(E) decreases very steeply with E.
- – At [math]T_2 = 600 K[/math], f(E) decreases less steeply, meaning more high-energy particles.
- ⇒ Practical Applications of the Boltzmann Factor
- 1. Population Ratios of Quantum States:
- The Boltzmann factor explains why high-energy states are less populated at low temperatures.
- At higher temperatures, higher energy states become more populated.
- 2. Chemical Reactions:
- The probability of a reaction depends on the number of molecules with energy greater than the activation energy ([math]E_a[/math]).
- The Boltzmann factor shows that higher temperatures increase the number of reactive molecules.
- 3. Phase Transitions:
- Processes like evaporation or sublimation rely on particles gaining enough energy to overcome intermolecular forces. The Boltzmann factor predicts how temperature affects this process.
c) Make calculations and estimates involving:
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I) Ratios of Characteristic Energies to the Energy and the Boltzmann Factor
- ⇒ Characteristic Energies and [math]kT[/math]
- The Boltzmann constant k relates temperature to energy. The characteristic thermal energy is given by:
- [math]kT[/math]
- Where:
- – [math]k = 1.38 × 10^{-23} J/K[/math] is the Boltzmann constant.
- – T is the absolute temperature in kelvins (K).
- – At room temperature (T≈300 K):
- – [math]kT = (1.38 × 10^{-23})(300) ≈ 4.14 × 10^{-21}[/math]
- This value is a useful benchmark for estimating how characteristic energies compare to [math]kT[/math].
- ⇒ Ratio of Energy to [math]kT : \frac{E}{kT}[/math]
- The ratio of characteristic energy (E) to thermal energy (kT) determines how likely particles are to occupy states with energy E. This ratio directly influences the Boltzmann factor.
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II) Boltzmann Factor:
- The Boltzmann factor is the probability of a particle occupying a state with energy E, relative to the probability of it being in the ground state. It is expressed as:
- [math]f(E) = e^{-\frac{E}{kT}}[/math]
- If [math]E ≪ kT[/math]:
- – [math]E/kT[/math] is small, so [math] e^{-\frac{E}{kT}} ≈ 1[/math].
- – High probability of particles occupying the state.
- If [math]E ≫ kT[[/math]:
- – [math]E/kT[/math] is large, so [math] e^{-\frac{E}{kT}} ≈ 0[/math].
- – Very low probability of particles occupying the state.
- ⇒ Example Calculations
- Example 1: Population Ratio Between Two States
- For two states with energies [math]E_1[/math] and [math]E_2[/math]:
- [math]N_2/N_1 = e^{-\frac{(E_2 – E_1)}{kT}}[/math]
- Suppose:
- – [math]E_1 = oJ[/math](ground state energy),
- – [math]E_2 = 2.0 10^{-20} J[/math]
- – [math]T = 300 K[/math],
- Calculate
- [math]\frac{E_2}{kT} = ? \\ \frac{E_2}{kT} = \frac{2.0 \times 10^{-20}}{4.14 \times 10^{-21}} \\
\frac{E_2}{kT} \approx 4.83 [/math] - Boltzmann factor:
- [math]\frac{N_2}{N_1} = e^{-\frac{E_2}{kT}} \\
\frac{N_2}{N_1} = e^{-4.83} \\
\frac{N_2}{N_1} \approx 0.008[/math] - Interpretation: At 300 K, the higher-energy state is 0.8% as likely to be occupied as the ground state.
- ⇒ Example 2: Activation Energy for a Reaction
- The rate of a chemical reaction depends on the fraction of molecules with energy greater than the activation energy ([math]E_a[/math]). This fraction is proportional to:
- [math]f(E_a) = e^{-\frac{E_a}{kT}}[/math]
- Suppose:
- – [math]E_a = 1 × 10^{-19} J[/math],
- – T=600K.
- Calculate [math]\frac{E_a}{kT}[/math]:
- [math]\frac{E_a}{kT} = \frac{1 \times 10^{-19}}{(1.38 \times 10^{-23}) \times 600} \\
\frac{E_a}{kT} = \frac{1 \times 10^{-19}}{8.28 \times 10^{-21}} \\
\frac{E_a}{kT} \approx 12.08[/math] - Boltzmann factor:
- [math]f(E_a) = e^{-\frac{E_a}{kT}} \\
f(E_a) = e^{-12.08} \\
f(E_a) \approx 5.63 \times 10^{-6}[/math] - Interpretation: At 600 K, only about [math]5.63 × 10^{-6}[/math] (or 0.000563%) of the molecules have enough energy to overcome the activation energy.
- ⇒ Example 3: Ratio for Ionization Energy
- The ionization energy of hydrogen is [math]E_ion = 2.18 × 10^{-18} J[/math]. At room temperature (T=300 K):
- [math]\frac{E_{\text{ion}}}{kT} = \frac{2.18 \times 10^{-18}}{4.14 \times 10^{-21}} \\
\frac{E_{\text{ion}}}{kT} \approx 526.57[/math] - Boltzmann factor:
- [math]f(E_{\text{ion}}) = e^{-\frac{E_{\text{ion}}}{kT}} \\
f(E_{\text{ion}}) = e^{-526.57} \\
f(E_{\text{ion}}) \approx 0[/math] - Interpretation: At room temperature, almost no hydrogen atoms have sufficient energy to ionize.
- ⇒ Practical Uses
- Quantum States:
- – Predict the populations of quantum energy levels in atoms or molecules.
- Chemical Reactions:
- – Determine how temperature affects reaction rates by examining the fraction of molecules that surpass activation energy.
- Material Science:
- – Analyze processes like thermionic emission, conduction in semiconductors, or phase transitions.