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A set of 13 particles occupies states with energies of \(0,100 .,\) and \(200 . \mathrm{cm}^{-1} .\) Calculate the total energy and number of microstates for the following energy configurations: a. \(a_{0}=8, a_{1}=5,\) and \(a_{2}=0\) b. \(a_{0}=9, a_{1}=3,\) and \(a_{2}=1\) \(\mathbf{c} . a_{0}=10, a_{1}=1,\) and \(a_{2}=2\) Do any of these configurations correspond to the Boltzmann distribution?

Short Answer

Expert verified
The total energy for all three configurations is \( 500 \mathrm{cm}^{-1} \), but the number of microstates for each configuration are \( 1287, 572,\) and \( 78 \) respectively. Therefore, none of these configurations correspond to the Boltzmann distribution due to differing microstates.

Step by step solution

01

Calculate the total energy for each configuration

We are given the number of particles in each energy state for the three configurations: a. \( a_0=8, a_1=5, \) and \( a_2=0 \) b. \( a_0=9, a_1=3, \) and \( a_2=1 \) c. \( a_0=10, a_1=1, \) and \( a_2=2 \) To calculate the total energy for each configuration, multiply the number of particles in each energy state by the corresponding energy and sum them up: a. Total energy = \( 0*a_0 + 100*a_1 + 200*a_2 = 0*8 + 100*5 + 200*0 = 500 \mathrm{cm}^{-1} \) b. Total energy = \( 0*a_0 + 100*a_1 + 200*a_2 = 0*9 + 100*3 + 200*1 = 500 \mathrm{cm}^{-1} \) c. Total energy = \( 0*a_0 + 100*a_1 + 200*a_2 = 0*10 + 100*1 + 200*2 = 500 \mathrm{cm}^{-1} \)
02

Calculate the number of microstates for each configuration

To calculate the number of microstates for each configuration, we will use the formula to distribute particles among energy states: Number of microstates = \( \frac{N!}{a_{0}!a_{1}!a_{2}!} \), where N is the total number of particles. Calculate the number of microstates for each configuration: a. Number of microstates = \( \frac{13!}{8!5!0!} = 1287 \) b. Number of microstates = \( \frac{13!}{9!3!1!} = 572 \) c. Number of microstates = \( \frac{13!}{10!1!2!} = 78 \)
03

Determine if each configuration corresponds to the Boltzmann distribution

To check if a given configuration corresponds to the Boltzmann distribution, we need to compare the calculated microstates and total energy. In this case, we can see that although all three configurations have the same total energy, they have different numbers of microstates, which means that not all configurations will correspond to the Boltzmann distribution. We can conclude that none of these configurations correspond to the Boltzmann distribution because the number of microstates is not equal in all three configurations for having the same total energy.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Boltzmann Distribution
The Boltzmann distribution is a cornerstone of statistical mechanics, and it describes the probability that a particle will be in a certain state as a function of the state's energy and the temperature of the system. According to this distribution, a particle is more likely to occupy a lower energy state than a higher energy state, which is a reflection of the natural tendency of systems to minimize their energy.

However, this doesn't mean that all particles will be found in the lowest energy state, as there is always thermal energy that allows particles to occupy higher energy states. The distribution is governed by the following equation:

\( P(E) = \frac{1}{Z} e^{-E/kT} \)

where \( P(E) \) is the probability of occupancy of the state with energy \( E \), \( Z \) is the partition function, \( k \) is the Boltzmann constant, and \( T \) is the temperature in kelvins. The partition function acts as a normalization factor ensuring that the probabilities for all possible states sum up to one.

The insight provided by the Boltzmann distribution is pivotal when considering different energy configurations of particles within a closed system. It allows us to predict not only which energy levels are most likely to be occupied but also to understand the relative population of different energy levels at equilibrium.
Statistical Mechanics
Statistical mechanics is the branch of physics that employs statistical methods to deal with large populations of particles and relates their microscopic properties to the macroscopic behaviors of the system. It bridges the gap between the laws of mechanics, which describe the behavior of individual particles, and the thermodynamic behavior of systems.

The core idea of statistical mechanics is that while the exact behavior of a massive number of particles is impossible to calculate individually, their average behavior can be described probabilistically. These average behaviors are linked to temperatures, pressure, volume, and other macroscopic properties observable in everyday life.

An integral aspect of statistical mechanics is the concept of microstates and macrostates. A macrostate is the macroscopic state of the system (like temperature or pressure), while a microstate represents the specific details of the system's constituent particles. A key principle of statistical mechanics is that macrostates with more corresponding microstates are more probable. The macrostate that maximizes the number of microstates is the system in equilibrium, and its properties can often be deduced by the methods of statistical mechanics.
Particle Distribution
Particle distribution pertains to the arrangement of particles among available energy states within a system. In the realm of statistical mechanics, it's crucial to understand how particles are distributed over different energy states, especially when analyzing systems at thermal equilibrium.

Each distribution is described by a series of microstates, which represent unique ways in which a system's particles can be arranged while preserving the overall macroscopic properties. Microstates are counted based on the principle of indistinguishability of particles, meaning particles are considered identical, and permutations among them do not count as separate microstates. The number of microstates associated with a particular energy distribution is a measure of the system's entropy.

The calculation of microstates is typically performed using the formula:

\( \text{Number of microstates} = \frac{N!}{\text{Product of each n}_i!} \)

where \( N \) is the total number of particles, and \( n_i \) is the number of particles in each energy state, represented as factorial terms in the denominator. For systems with distinguishable particles, this formula changes to accommodate the different counting statistics.

Understanding particle distribution and the corresponding number of microstates provides insights into the system's degree of disorder or entropy and is instrumental in predicting the likelihood of different macrostates based on the principles of statistical mechanics.

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Most popular questions from this chapter

Determine the partition function for the vibrational degrees of freedom of \(\mathrm{Cl}_{2}\left(\widetilde{\nu}=525 \mathrm{cm}^{-1}\right)\) and calculate the probability of occupying the first excited vibrational level at 300 . and \(1000 .\) K. Determine the temperature at which identical probabilities will be observed for \(\mathrm{F}_{2}\left(\widetilde{v}=917 \mathrm{cm}^{-1}\right)\).

Problem numbers in red indicate that the solution to the problem is given in the Student's Solutions Manual.a. What is the possible number of microstates associated with tossing a coin \(N\) times and having it come up \(H\) times heads and \(T\) times tails? b. For a series of 1000 tosses, what is the total number of microstates associated with \(50 \%\) heads and \(50 \%\) tails? c. How much less probable is the outcome that the coin will land \(40 \%\) heads and \(60 \%\) tails?

In Example Problem 30.1 , the weights associated with observing 40 heads and 50 heads after flipping a coin 100 times were determined. Perform a similar calculation to determine the weights associated with observing 400 and 500 heads after tossing a coin 1000 times. (Note: Stirling's approximation will be useful in performing these calculations).

The probability of occupying a given excited state \(p_{i}\) is given by \(p_{i}=n_{i} / N=e^{-\beta \kappa_{i}} / q,\) where \(n_{i}\) is the occupation number for the state of interest, \(N\) is the number of particles, and \(\varepsilon_{i}\) is the energy of the level of interest. Demonstrate that the preceding expression is independent of the definition of energy for the lowest state.

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