Derivative Maxwell boltzmann distribution

In summary, the goal is to show that the peak of the Maxwell Boltzmann distribution is equal to 1/2kt. This involves taking the derivative of the distribution function and setting it equal to 0. After some simplification, the derivative is found to be 1/2kT=E^1/4. By solving for E, the value of E that maximizes N(E) can be found.
  • #1
giraffe
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0

Homework Statement


i need to show that the peak of the maxwell Boltzmann distribution is equal to 1/2 kt.

Homework Equations


maxwell Boltzmann distribution according to modern physics 3rd edition by kenneth kramer.

ill try to do my best with this

[itex] N(E)= \frac{2N}{√∏} \frac{1}{(kT)^\frac{3}{2}} E^\frac{1}{2} e^\frac{-E}{kT}[/itex]

N is the total number of molecules while N(E) is the distribution function (with units energy to the -1) defined so that N(E) dE is the number of molecules dN in the energy interval dE at E. dn=N(E)dE

The Attempt at a Solution



so i need to take the derivative and set that equal to 0 and hope i get 1/2kt. I am having trouble with the derivative itself. I am taking the derivative with respect to E so everything else is considered a constant. so to try to make this easier i took all that junk in front of the E and said it is just some constant a. that allowed me to go through and do the product rule. after that, I've been trying to simplify it but am getting nowhere. need some advice on how to do this properly as i believe I am not.

thanks a bunch
 
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  • #2
Hi giraffe! Do you think you could show some more work so that we can see where exactly you are getting stuck?
 
  • #3
i can try.

as i mentioned [itex] a = \frac{2N}{√∏} \frac{1}{(kT)^\frac{3}{2}} [/itex] which is just a constant since we are doing this with respect to E

so now you have [itex] (aE^ \frac{1}{2})(e^\frac{-E}{kT}) [/itex]

follow the product rule: derivative of first * second + first *derivative of second

[itex] \frac{aE^ \frac{-1}{2} e^ \frac{-E}{kT}}{2} - \frac{aE^ \frac{1}{2} e^ \frac{-E}{kT}}{kT} [/itex]

after that I've tried to combine the fractions and cancel some stuff out but nothing works.
 
  • #4
Good job so far. Now remember that ##\frac{aE^ \frac{-1}{2} e^ \frac{-E}{kT}}{2} - \frac{aE^ \frac{1}{2} e^ \frac{-E}{kT}}{kT}=0## since we are talking about maximizing (the derivative must equal zero). Do you think you could do some cancellations to simplify?
 
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  • #5
ahhh...silly me...thanks...if i add the 2nd term over then the numerators will cancel out leaving 1/2 on left and 1/kT on right. multiply by kT and you have the 1/2kT=0 which is what i want.

thanks again for helping me with that simple mistake.
 
  • #6
giraffe said:
ahhh...silly me...thanks...if i add the 2nd term over then the numerators will cancel out leaving 1/2 on left and 1/kT on right. multiply by kT and you have the 1/2kT=0 which is what i want.

thanks again for helping me with that simple mistake.

Whoa! The E's don't cancel since there's ##E^{-1/2}## on the left and ##E^{1/2}## on the right.
 
  • #7
Correct. Rearranging you'll get 1/2kT=E^1/4 which is the final derivative. Set E=0 to find max.
 
  • #8
Not correct. When you multiply both sides by ##E^{1/2}## you add the exponents of ##E##, not multiply them (which is what I think you did). If you then solve for ##E##, you find the value of ##E## that maximizes ##N(E)##. You don't have to take another derivative.
 

What is the Derivative Maxwell Boltzmann Distribution?

The Derivative Maxwell Boltzmann Distribution is a probability distribution that describes the distribution of velocities of particles in a gas at a given temperature. It is a derivative of the original Maxwell Boltzmann Distribution, which describes the distribution of speeds of particles in a gas.

What is the significance of the Derivative Maxwell Boltzmann Distribution?

The Derivative Maxwell Boltzmann Distribution is significant because it allows us to calculate the number of particles with a specific velocity in a gas at a given temperature. This is important in understanding the behavior of gases and their properties.

How is the Derivative Maxwell Boltzmann Distribution calculated?

The Derivative Maxwell Boltzmann Distribution is calculated using the following formula: f(v) = (m/2πkT)^1/2 * v^2 * e^(-mv^2/2kT), where m is the mass of the particle, k is the Boltzmann constant, T is the temperature, and v is the velocity of the particle.

What is the relationship between temperature and the Derivative Maxwell Boltzmann Distribution?

The temperature of a gas directly affects the shape of the Derivative Maxwell Boltzmann Distribution curve. As the temperature increases, the curve becomes wider and flatter, indicating a greater spread of velocities in the gas. As the temperature decreases, the curve becomes narrower and taller, indicating a smaller spread of velocities.

What are some applications of the Derivative Maxwell Boltzmann Distribution?

The Derivative Maxwell Boltzmann Distribution is used in various fields, including physics, chemistry, and engineering. It is important in understanding the flow of gases in pipes, the behavior of molecular gases, and the design of gas turbines and other devices that use gases. It is also used in astrophysics to study the dynamics of gases in space.

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