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

[Drakkith]

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!

 

[Delphi51]

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.

 

[fzero]

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 $$c$$.

 

[Drakkith]

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?

 

[Delphi51]

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

 

[Dickfore]

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:

$$E_{0} = m c^{2}$$

and the formula is:

$$v = \sqrt{\frac{2 K}{m}} = \sqrt{\frac{2 K c^{2}}{m c^{2}}} = c \, \sqrt{\frac{2 K}{m c^{2}}}$$

Here K is the kinetic energy.