What is the formula for finding the speed of ions in a heated plasma?

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SUMMARY

The discussion focuses on calculating the velocity of positive ions in a deuterium plasma heated to 15 keV. The correct approach involves using the kinetic energy formula, v = sqrt(2E/m), where E is the energy in joules and m is the mass of deuterium, approximately 3.3444 x 10-27 kg. Converting 15 keV to joules results in 2.4032647305 x 10-15 J, leading to a calculated ion velocity of roughly 1 million m/s. The discussion emphasizes the importance of maintaining consistent units throughout the calculations.

PREREQUISITES
  • Understanding of kinetic energy equations, specifically v = sqrt(2E/m)
  • Knowledge of energy conversion from electronvolts (eV) to joules (J)
  • Familiarity with the mass of deuterium, approximately 3.3444 x 10-27 kg
  • Basic principles of plasma physics and ion behavior
NEXT STEPS
  • Research the conversion of energy units, specifically from eV to J
  • Learn about the properties of deuterium and its applications in plasma physics
  • Explore advanced kinetic theory and its implications in plasma behavior
  • Study the relativistic effects on ion velocities in high-energy plasmas
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Physicists, plasma researchers, and students studying plasma dynamics or thermonuclear fusion who require a deeper understanding of ion behavior in heated plasmas.

Drakkith
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First off, this isn't a homework problem, but something I am trying to figure out on my own. I figured there wasn't a better place in the forums to ask though.

Homework Statement


I'm trying to find the velocity of positive ions in a deuterium plasma that is heated to 15keV.


Homework Equations


I'm not sure of the correct ones to use. I've tried the following.
1 eV = 1.602176487(40)×10−19 J
Ke=1/2MV*2


The Attempt at a Solution


I've found the mass of dueterium as about 3.3444 x 10*-27 Kg.
Converting 15keV to J gives me 24032.647305 x 10*-19 J.
Plugging this into the kinetic energy formula doesn't work very well, as i get a V of less than 1. Am I not converting 15keV to the right amount of joules? I don't know if the temp in eV would convert directly to Joules.
Thanks!
 
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Better check that calc. Doing v = sqrt(2E/m) you'll have 10^-15 in the numerator and 10^-27 in the denominator. That gives the square root of 10^12 or roughly 1 million m/s for the velocity.
 
Delphi51 said:
Better check that calc. Doing v = sqrt(2E/m) you'll have 10^-15 in the numerator and 10^-27 in the denominator. That gives the square root of 10^12 or roughly 1 million m/s for the velocity.

Just to make an observation. If you use v = \sqrt{2E/m}, you can leave everything in eV to obtain the speed in units of c.
 
Delphi51 said:
Better check that calc. Doing v = sqrt(2E/m) you'll have 10^-15 in the numerator and 10^-27 in the denominator. That gives the square root of 10^12 or roughly 1 million m/s for the velocity.

It's been a long time since I did any serious math. Whats the rules again when you have to mulitply and divide with unlike exponents?
 
To divide powers with the same base, keep the same base and subtract the exponents.
10^3 divided by 10^-2 is 10^(3 - -2) = 10^(3+2) = 10^5
 
fzero said:
Just to make an observation. If you use v = \sqrt{2E/m}, you can leave everything in eV to obtain the speed in units of c.

yes, since when you report the mass in energy units you refer to the rest energy of the particle:

<br /> E_{0} = m c^{2}<br />

and the formula is:

<br /> v = \sqrt{\frac{2 K}{m}} = \sqrt{\frac{2 K c^{2}}{m c^{2}}} = c \, \sqrt{\frac{2 K}{m c^{2}}}<br />

Here K is the kinetic energy.
 

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