Maxwell-Boltzmann Distribution

In summary, we discussed a collection of N noninteracting atoms with a single excited state at energy E, assumed to obey Maxwell-Boltzmann statistics with nondegenerate ground and excited states. We calculated the ratio of excited to ground state atoms, the average energy of an atom, the total energy of the system, and the heat capacity. We also explored the relationship between the probability and energy in this system, and how it can be used to calculate the average energy.
  • #1
mkg0070
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1. Consider a collection of N noninteracting atoms with a single excited state at energy E. Assume the atoms obey the Maxwell-Boltzmann statistics, and take both the ground state and the excited state to be nondegenerate. a.) At temperature T, what is the ratio of the number of atoms in the excited state to the number in the ground state? b.) What is the average energy of an atom in this system? c.) What is the total energy of this system? d.) What is the heat capacity of this system?



2. f(E)=(A^-1)*[e^(-E/kT)]
p(E)=g(E)f(E)
p(E2)/P(E1)=[g(E2)/g(E1)]*{e^[-(E2-E1)/kT]}



3. The answer to part "a" is: e^(-E/kT) which I understand since both the ground state and the excited states are nondegenerate.
The answer to part "b" is: E/[1+e^(E/kT)]. This is the part I do not understand. Can you please explain why this is the answer and how to get this?
 
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  • #2
mkg0070 said:
1. Consider a collection of N noninteracting atoms with a single excited state at energy E. Assume the atoms obey the Maxwell-Boltzmann statistics, and take both the ground state and the excited state to be nondegenerate. a.) At temperature T, what is the ratio of the number of atoms in the excited state to the number in the ground state? b.) What is the average energy of an atom in this system? c.) What is the total energy of this system? d.) What is the heat capacity of this system?



2. f(E)=(A^-1)*[e^(-E/kT)]
p(E)=g(E)f(E)
p(E2)/P(E1)=[g(E2)/g(E1)]*{e^[-(E2-E1)/kT]}



3. The answer to part "a" is: e^(-E/kT) which I understand since both the ground state and the excited states are nondegenerate.
The answer to part "b" is: E/[1+e^(E/kT)]. This is the part I do not understand. Can you please explain why this is the answer and how to get this?



the average energy is equal to

[tex] E_0 P(E_0) + E_1 P(E_1) [/tex]

where the probability of obtaining each energy is simply

[tex] P(E_i) = \frac{e^{-E_i /kT}}{e^{-E_0/kT} + e^{-E_1/kT}} [/tex]

Try this
 

What is the Maxwell-Boltzmann Distribution?

The Maxwell-Boltzmann Distribution is a probability distribution that describes the speed of molecules in a gas at a given temperature. It is named after James Clerk Maxwell and Ludwig Boltzmann who developed the theory in the 19th century.

What does the Maxwell-Boltzmann Distribution tell us about gas molecules?

The distribution tells us the likelihood of finding molecules with a certain speed in a gas at a given temperature. It shows that most molecules have speeds close to the average, and the number of molecules decreases as the speed increases.

How is the Maxwell-Boltzmann Distribution related to the Kinetic Theory of Gases?

The Kinetic Theory of Gases explains the behavior of gases at a microscopic level, while the Maxwell-Boltzmann Distribution provides a mathematical representation of this behavior. The distribution is derived from the assumptions of the Kinetic Theory, such as the random motion of molecules and collisions between molecules.

What factors affect the shape of the Maxwell-Boltzmann Distribution?

The shape of the distribution is affected by temperature, molecular mass, and the type of gas. Higher temperatures result in a broader distribution with a higher average speed, while lighter molecules and gases with fewer degrees of freedom have narrower distributions with higher peak speeds.

What practical applications does the Maxwell-Boltzmann Distribution have?

The distribution is used in various fields, such as thermodynamics, statistical mechanics, and chemical kinetics, to calculate the properties of gases. It also helps in understanding and predicting phenomena such as diffusion, evaporation, and gas reactions.

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