Percentage error in Maxwellian distribution.

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The discussion centers on calculating the average kinetic energy of a molecule following Maxwellian distribution, using the formula ½mc², where c is the average velocity. The average velocity is derived from Maxwell's velocity distribution, resulting in c' = √(8kT/πm). Participants inquire about calculating percentage error and the standard deviation, with one contributor expressing difficulty in understanding the statistical concepts involved. Suggestions are made to look up standard deviation calculations, indicating a need for clearer explanations. The conversation highlights the connection between kinetic energy calculations and statistical methods in physics.
debjit625
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Homework Statement


In computing the average kinetic energy of a molecule obeying Maxwellian distribution one use the formula ½mc2 .Calculate the percentage error incurred in the calculation.

The Attempt at a Solution


Here c is the average velocity of a molecule obeying Maxwellian distribution which I can calculate which is √(8KT/πm), but I have no idea how to calculate the percentage error so no attempt has been made.

Thanks
 
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Hello debjit,

What equation did you use to calculate the average kinetic energy ?
Do you know the equation for the standard deviation ?
 
Thanks for the reply BvU.
I used Maxwell's velocity distribution to compute average or mean speed

Maxwell's velocity distribution
dNc = 4πN(m/2πkT)3/2e-(mc2/2kT)c2dc

From this I computed the average or mean speed
c' = 1/N ( o cdNc)

c' = √(8kT/πm)

"Do you know the equation for the standard deviation ?"
Well I am not very much good at the statistical part ,so if you kindly explain it it will be very helpful.

Thanks
 
debjit625 said:
Well I am not very much good at the statistical part ,so if you kindly explain it it will be very helpful
Well, you can always look it up ...
usually you do something like ##<x^2> - <x>^2##, so you're halfway already :smile:
 

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