# Kinetic energy of an electron inside the sun [closed]

I have to tell if an average proton or electron at 15 million degrees kelvin be treated by relativistic mechanics. The criterion is that the kinetic energy differ from the classical kinetic energy by more than 1%. The average kinetic energy of a particle at temperature T is $$\frac{3}{2}kT$$, where k is Boltzmann's constant.

Since T relativistic = $$\frac{3}{2}kT$$ and $$E = \gamma mc^2 = mc^2 + T$$

I have $$\gamma mc^2 = mc^2 +\frac{3}{2}kT$$

And from the formula above I get $$\gamma = \frac{3KT}{2mc^2} + 1$$

From what I know $$mc^2 = 0.5MeV = 500000eV$$ for an electron

Plugging the the numbers I have $$\gamma = 6.21^{-22} + 1$$

$$\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = 6.21^{-22}$$ to find $$v$$

However, I still can't compare the kinectic energies since I don't have $$m$$. For example $$\frac{\frac{1}{2}mv^2}{\frac{3}{2}KT}$$

Furthermore, I have some weird values for the velocity. Even $$\gamma$$ sounds not right. I don't thing that the velocity of an electron at this temperature should be that low. I have $$v = (3.52 \cdot 10^-8)c$$

• It is obvious you are mixing units. What do you mean you don't know the mass? It's just the mass of an electron or proton, whichever you are considering. You know, or can look up all the values in your equation in SI units. There is no need to use MeV. Just plug and chug to solve for gamma, or the velocity v. Nov 16 '21 at 1:10
• I'm guessing I can't use the mass of the electron since I don't have it in the statement. $mc^2$ should be 0.5Mev, no? Nov 16 '21 at 1:15
• If you have k in joule/kelvin, then $mc^2$ must be in joules. Nov 16 '21 at 1:21
• You are right, but even if plug the mass in kg I get a velocity close to c. I don't thing the electron should move at the speed of light. Nov 16 '21 at 1:25
• I get v/c around 0.05 for an electron. Nov 16 '21 at 1:35

$$\frac{e}{k}\times (1\,\rm V) = 11604.5\,{\rm J/eV(J/K)} \approx 12000 \,{\rm K/eV}$$
Thus $$15000000\,K$$ is $$KE=1250\,$$ eV (with $$c=1$$). Compare that with $$m_e = 511\,{\rm keV}$$ to get a Lorentz factor:
$$\gamma = \frac{KE+m}m=\frac{512250}{511000} \approx 1.0024$$