Fermions that can access 10 distinct energy states; Statistical Physics

In summary: The entropy of the system is basically the logarithm of the number of different energy levels that the system can occupy. In this case, there are 10 energy levels, so the entropy is 10 log(10) = 1.0.
  • #1
matt_crouch
161
1

Homework Statement



Consider a system made of 4 quantum fermions that can access 10 distinct states respectively with energies:

En=n/10 eV with n=1,2,3,4,5,6,7,8,9,10

1) Write the expression for the entropy when the particles can access all states with equal probability

2) Compute the Entropy of the isolated system at energy U =1 eV

3) Compute the entropy of the isolated system at energy 1.1 eV

Homework Equations



Ω=G!/m!(G-m)!
s=kBln(Ω)


The Attempt at a Solution



the first question i think is answered basically by the first equation i gave for the statistical weight because that is for indistinguishable particles with multiple occupancy not allowed. I am a little bit stuck on the 2nd and 3rd questions. The probability of finding a particle in the lowest state must be more probable than finding a particle in the highest state but the equation for the statistical weight won't take that into account. if i can be pointed in the right direction that would be awesome :smile:
 
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  • #2
matt_crouch said:
the first question i think is answered basically by the first equation i gave for the statistical weight because that is for indistinguishable particles with multiple occupancy not allowed. I am a little bit stuck on the 2nd and 3rd questions. The probability of finding a particle in the lowest state must be more probable than finding a particle in the highest state but the equation for the statistical weight won't take that into account. if i can be pointed in the right direction that would be awesome :smile:

The physical way of looking at entropy is that it's the logarithm of the number of states corresponding to a given energy. For example, when the total energy is fixed to 1eV, there's only one possible arrangement to get this: 1/10eV + 2/10eV + 3/10eV + 4/10eV = 1eV. Now it becomes just a combinatorics problem, and you only need to figure out how many such configurations there are.
 
  • #3
ok so because there is only 1 possible arrangement for u=1ev the statistical weight can be calculated used the equation above so

Ω= 4!/4!(1)!
Ω=1

so s=kln(1)

then for the u=1.1 ev so the only possible state will be when 3/10 + 5/10 + 2/10 + 1/10 = 1.1 ev

so again Ω=1

s=kln (1)
 
  • #4
Oops, I forgot about spin. If your fermions have spin 1/2, you can have two of them occupying a state with same n. Maybe you should at least mention this, if not calculate it completely.
 
  • #5


1) The expression for the entropy when the particles can access all states with equal probability can be written as:

S = k_B ln(Ω)

where k_B is the Boltzmann constant and Ω is the number of microstates available to the system. In this case, since all states are equally probable, the number of microstates can be calculated using the equation:

Ω = (G!)/((m!)(G-m)!)

where G is the total number of states (in this case, 10) and m is the number of particles (in this case, 4). Plugging in these values, we get:

Ω = (10!)/((4!)(10-4)!) = 210

2) To compute the entropy of the isolated system at energy U = 1 eV, we need to first calculate the number of particles in each energy state. This can be done using the Fermi-Dirac distribution:

n(E) = 1/(e^((E-E_F)/k_B T) + 1)

where E_F is the Fermi energy and k_B is the Boltzmann constant. In this case, since we are dealing with fermions, the Fermi energy is equal to the highest occupied energy state, which in this case is En=4/10 eV. Plugging in these values, we get:

n(1 eV) = 1/(e^((1 eV - 4/10 eV)/(k_B T)) + 1) = 1/(e^(6/10 k_B T) + 1)

Now, we can use this value to calculate the entropy using the expression from part 1:

S = k_B ln(Ω) = k_B ln(210) = 5.34 k_B

3) To compute the entropy of the isolated system at energy 1.1 eV, we can follow a similar approach as in part 2. First, we calculate the number of particles in each energy state using the Fermi-Dirac distribution:

n(1.1 eV) = 1/(e^((1.1 eV - 4/10 eV)/(k_B T)) + 1) = 1/(e^(7/10 k_B T) + 1)

Then, we can use this value to calculate the entropy using the expression from part
 

1. What are fermions?

Fermions are a type of elementary particle that make up the building blocks of matter. They include particles such as electrons, protons, and neutrons.

2. How many energy states can fermions access?

Fermions can access 10 distinct energy states in this particular scenario. However, this number can vary depending on the specific system being studied.

3. What is Statistical Physics?

Statistical Physics is a branch of physics that uses statistical methods to study the behavior of large systems of particles. It aims to understand macroscopic properties of matter through the behavior of individual particles.

4. How are fermions and energy states related in Statistical Physics?

In Statistical Physics, fermions are studied in terms of their energy states. The energy states of fermions determine their behavior and characteristics within a given system.

5. Can fermions access an infinite number of energy states?

No, fermions cannot access an infinite number of energy states. The number of energy states available to fermions is limited by the specific system they are in and the energy levels within that system.

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