# Finding the Speed of Ions

Staff Emeritus
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 in to 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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Delphi51
Homework Helper
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
Homework Helper
Gold Member
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 $$c$$.

Staff Emeritus
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
Homework Helper
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

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.