Calculate most probable energy from Maxwell-Boltzmann distribution of energy

In summary, the problem is asking for the most probable energy and integrating from 0 to infinity. Some kind of integration is needed, but it is not a gamma function. The most probable energy is found by setting the derivative to 0 and solving for E.
  • #1
TehDarkArchon
18
0

Homework Statement


If the Maxwell-Boltzmann distribution of energy is f(E)=2*pi*E1/2*(1/pi*k*T)3/2*eE/kT. Can you calculate the most probable energy from this? (The answer is kt/2).


Homework Equations


Some kind of integration (guassian integral) is needed from 0 to infinity I believe.


The Attempt at a Solution


I tried integration by parts and some other methods with no luck. My teacher made it clear that this is not a gamma function (my initial thought since the E1/2 is paired with an exponential function) and that the answer is relatively easy. Any help would be much appreciated!
 
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  • #2
Wikipedia says that it is a gamma distribution...
Also, shouldn't it be e-E/kT?
 
  • #3
It is supposed to be e^-E/kT, sorry about that. Unfortunately he made it clear that it's not a gamma function...and integration by parts only causes that term to be 1/2(E)^1/2 and just ends up being more complicated. We have discussed transformation of variables and guassian integrals in class, so perhaps that's the way to go about this problem (I'm still getting used to both of these as I've never done them before this class, thus probably why I'm struggling so much.)
 
  • #4
You don't need to integrate at all. You're thinking of the average value, or expectation value, of energy. The problem is asking for the most probable energy.
 
  • #5
Okay that makes sense...I was going off my teacher's example of how to calculate most probable velocity from the Maxwell-Boltzmann distribution. Any hint on what kind of calculation I need to do to get most probable energy? I'm probably going to feel incredibly stupid once I figure it out...
 
  • #6
What does the phrase "most probable" translate to mathematically in terms of f(E)?
 
  • #7
Think about the curve of f(E) against E.
 
  • #8
f(E) is just a function of E, with the most probable energy being the energy that most of the molecules have in the system, so should I just compute the value to infinity?
 
  • #9
No. What does f(E) represent?
 
  • #10
I know we already have some people helping but, think "optimization problems"...
 
  • #11
Yep, jfy4 has the right idea.
Another hint: If you had the graph of f(E) against E, you could simply see by looking at it what the most probable energy was.
 
  • #12
Ahh I think I see how to get it now. Basically Emp is the maximum of f(E) so you'd go about this problem how you would any graph to find the maximum, by setting the derivative to 0. By setting all of the constants equal to Q, canceling Q and the exponential function on both sides, i made [itex]\frac{1}{2\sqrt{E}}[/itex] = [itex]\frac{\sqrt{E}}{kT}[/itex]. With some rearrangement I get kT = 2E, or kt/2 = E.
 
  • #13
well done.
 
  • #14
Thanks very much everyone for your help! I appreciate it so much!
 

1. What is the Maxwell-Boltzmann distribution of energy?

The Maxwell-Boltzmann distribution of energy is a probability distribution that describes the distribution of energy among particles in a gas at a given temperature. It is named after James Clerk Maxwell and Ludwig Boltzmann, who developed the concept in the late 19th century.

2. How is the most probable energy calculated from the Maxwell-Boltzmann distribution?

The most probable energy is calculated by finding the value of energy at which the Maxwell-Boltzmann distribution has its peak. This can be done by taking the derivative of the distribution and setting it equal to zero. The resulting energy value is the most probable energy.

3. What factors affect the most probable energy in the Maxwell-Boltzmann distribution?

The most probable energy in the Maxwell-Boltzmann distribution is affected by the temperature of the gas and the mass of the particles. As temperature increases, the most probable energy also increases. Similarly, as particle mass increases, the most probable energy decreases.

4. Can the most probable energy be greater than the average energy in the Maxwell-Boltzmann distribution?

Yes, the most probable energy can be greater than the average energy in the Maxwell-Boltzmann distribution. This is because the average energy takes into account all possible energy values, while the most probable energy only considers the energy value at which the distribution has its peak.

5. How is the Maxwell-Boltzmann distribution of energy used in physics and chemistry?

The Maxwell-Boltzmann distribution of energy is used to understand the behavior of particles in a gas and to predict various properties such as pressure, temperature, and kinetic energy. It is also used in statistical mechanics to describe the distribution of energies in a system and in quantum mechanics to explain the behavior of particles at the atomic level.

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