# Converting 1 Kelvin into eV

I am using $E=\frac{3}{2} kT$ to find the equivalent of 1 K in eV using $k=1.38\times 10^{-23} \,\mathrm{\frac{J}{K}}$. I get $E= 1.29 \times 10^{-4} \,\mathrm{eV}$, but Wikipedia says it's $8.6\times 10^{-5} \,\mathrm{eV}$. Could anyone tell me what Wikipedia did different then me?

• You got the units wrong, Kelvin is for temperature, and eV is for energy. Both represent two different things, – HyperBean Jul 23 '17 at 21:03
• The equation you're using is not a conversion. $E=\frac{3}{2}kT$ is the average kinetic energy of an ideal gas. – Johnathan Gross Jul 23 '17 at 21:06
• Thank you HyperBean and Johnathan Gross, it makes more sense now. – Rocket Hack Jul 23 '17 at 21:14
• @HyperBean There is a tight coupling between mean energies per (accessible, quadratic) mode in thermal systems and temperature, a fact that is encapsulated in Boltzmann's constant. – dmckee Jul 23 '17 at 21:23

It is easier to see when you use k in eV per K which is $$8.6173303(50)×10^{-5} \ \mathrm{\frac{eV}{K}}$$