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Consider the following energy levels, each capable of holding two particles: $$\begin{array}{l}{E=2 \mathrm{kJ}} \\ {E=1 \mathrm{kJ}} \\ {E=0 \quad \underline{X X}}\end{array}$$ Draw all the possible arrangements of the two identical particles (represented by X) in the three energy levels. What total energy is most likely, that is, occurs the greatest number of times? Assume that the particles are indistinguishable from each other.

Short Answer

Expert verified
We have six possible arrangements of the two identical particles in the three energy levels: \(XX \ \_ \ \_\), \(\_ \ XX \ \_\), \(\_ \ \_ \ XX\), \(X \ X \ \_\), \(X \ \_ \ X\), and \(\_ \ X \ X\). The corresponding total energies are 0 kJ, 2 kJ, 4 kJ, 1 kJ, 2 kJ, and 3 kJ. The most likely total energies are 1 kJ and 2 kJ, as they occur two times each.

Step by step solution

01

Understanding the problem statement

We are given three energy levels, each capable of holding two particles, and we have two distinguishable particles represented by X. The goal is to find all the possible arrangements of the two particles in the three energy levels.
02

Enumerate all possible arrangements

Let's start with the possible arrangements for placing the two particles in the energy levels. Remember, each energy level can hold up to two particles, and the particles are indistinguishable. 1. Both particles in the 0 kJ energy level: \(XX \ \_ \ \_\) 2. Both particles in the 1 kJ energy level: \(\_ \ XX \ \_\) 3. Both particles in the 2 kJ energy level \(\_ \ \_ \ XX\) 4. One particle in 0 kJ and another in 1 kJ energy level: \(X \ X \ \_\) 5. One particle in 0 kJ and another in 2 kJ energy level: \(X \ \_ \ X\) 6. One particle in 1 kJ and another in 2 kJ energy level: \(\_ \ X \ X\)
03

Calculate the total energy for each arrangement

Now let's calculate the total energy for each arrangement: 1. Both particles in the 0 kJ energy level: Total energy = 0 kJ 2. Both particles in the 1 kJ energy level: Total energy = 1 kJ + 1 kJ = 2 kJ 3. Both particles in the 2 kJ energy level Total energy = 2 kJ + 2 kJ = 4 kJ 4. One particle in 0 kJ and another in 1 kJ energy level: Total energy = 0 kJ + 1 kJ = 1 kJ 5. One particle in 0 kJ and another in 2 kJ energy level: Total energy = 0 kJ + 2 kJ = 2 kJ 6. One particle in 1 kJ and another in 2 kJ energy level: Total energy = 1 kJ + 2 kJ = 3 kJ
04

Find the total energy that occurs the most

Now that we have the total energy for each arrangement, let's count how many times each energy value appears: - 0 kJ occurs 1 time - 1 kJ occurs 2 times - 2 kJ occurs 2 times - 3 kJ occurs 1 time - 4 kJ occurs 1 time Since the total energy of 1 kJ and 2 kJ occur the greatest number of times (2 times each), they are the most likely total energy values.

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Key Concepts

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

Indistinguishable Particles
When we talk about indistinguishable particles, we mean that no matter how we arrange them, they look identical. This concept is crucial in many areas of physics, especially in quantum mechanics. In our problem, the particles are represented by X, and distinguishing between different Xs is impossible. Whether one X is placed in one energy level and the other in another, or vice versa, these setups are the same.

Understanding indistinguishable particles helps in simplifying complex systems. It allows us to count only unique configurations rather than each possible permutation, making calculations more straightforward. This distinction is essential for accurately determining energy and statistical properties.
Energy Calculations
In analyzing particle arrangements, we need to calculate the energy levels associated with each arrangement. These calculations are straightforward once you know where each particle is placed. Each particle contributes an energy corresponding to its level, and we simply add these values.

Here's how it's done:
  • If both particles are in the 0 kJ energy level, the total energy is 0 kJ.
  • If they're both in the 1 kJ level, it's 2 kJ (1 kJ per particle).
  • If both are in the 2 kJ level, it's 4 kJ (2 kJ per particle).
  • When particles are in different levels, like one in 0 kJ and the other in 1 kJ, the total energy is their sum, 1 kJ.

These calculations help evaluate which energy states are most probable.
Energy Level Diagrams
Energy level diagrams provide a visual representation of where particles are located across different energy levels. This diagram is a valuable tool for conceptualizing the distribution of particles. In our scenario, we have three energy levels: 0 kJ, 1 kJ, and 2 kJ, each capable of holding two particles.

An energy level diagram helps identify possible configurations. For instance, the arrangement \(XX \ \_ \ \_\) indicates both particles at 0 kJ while \(_ \ XX \ \_\) shows both at 1 kJ. This visualization aids in understanding how particles distribute themselves in various states and allows us to compute the total energy for each configuration effortlessly.
Probability in Energy Distribution
Probability in energy distribution helps determine which energy states are more likely to occur. It's about finding how often a particular total energy appears among all possible arrangements.

In our example, 1 kJ and 2 kJ appeared twice, making them the most probable energy levels. To determine these probabilities, count the number of configurations leading to each possible total energy.

For instance:
  • 0 kJ occurs once.
  • 1 kJ appears in two setups.
  • Similarly, 2 kJ appears twice too.
  • 3 kJ and 4 kJ each occur once.

Understanding this probability provides insights into the behavior of particle systems, which is vital for predicting system behavior in various physical conditions.

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Most popular questions from this chapter

What information can be determined from \(\Delta G\) for a reaction? Does one get the same information from \(\Delta G^{\circ},\) the standard free energy change? \(\Delta G^{\circ}\) allows determination of the equilibrium constant \(K\) for a reaction. How? How can one estimate the value of \(K\) at temperatures other than \(25^{\circ} \mathrm{C}\) for a reaction? How can one estimate the temperature where \(K=1\) for a reaction? Do all reactions have a specific temperature where \(K=1 ?\)

Human DNA contains almost twice as much information as is needed to code for all the substances produced in the body. Likewise, the digital data sent from Voyager II contained one redundant bit out of every two bits of information. The Hubble space telescope transmits three redundant bits for every bit of information. How is entropy related to the transmission of information? What do you think is accomplished by having so many redundant bits of information in both DNA and the space probes?

At 1 atm, liquid water is heated above \(100^{\circ} \mathrm{C}\) . For this process, which of the following choices (i-iv) is correct for \(\Delta S_{\text { surr }} ? \Delta S ?\) \(\Delta S_{\text { univ }} ?\) Explain each answer. i. greater than zero ii. less than zero iii. equal to zero iv. cannot be determined

The equilibrium constant K for the reaction $$2 \mathrm{Cl}(g) \rightleftharpoons \mathrm{Cl}_{2}(g)$$ was measured as a function of temperature (Kelvin). A graph of \(\ln (K)\) versus 1\(/ T\) for this reaction gives a straight line with a slope of \(1.352 \times 10^{4} \mathrm{K}\) and a y-intercept of \(-14.51 .\) Determine the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. See Exercise \(85 .\)

For each of the following pairs, which substance has the greater value of S? a. \(\mathrm{N}_{2} \mathrm{O}(\text { at } 0 \mathrm{K})\) or He (at 10 \(\mathrm{K} )\) b. \(\mathrm{N}_{2} \mathrm{O}(g)\) (at \(1 \mathrm{atm}, 25^{\circ} \mathrm{C} )\) or He(g) (at 1 atm, \(25^{\circ} \mathrm{C} )\) c. \(\mathrm{NH}_{3}(s)\) (at 196 \(\mathrm{K} ) \longrightarrow \mathrm{NH}_{3}(l)(\text { at } 196 \mathrm{K})\)

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