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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?

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    $\begingroup$ You got the units wrong, Kelvin is for temperature, and eV is for energy. Both represent two different things, $\endgroup$ – HyperBean Jul 23 '17 at 21:03
  • $\begingroup$ The equation you're using is not a conversion. $E=\frac{3}{2}kT$ is the average kinetic energy of an ideal gas. $\endgroup$ – Johnathan Gross Jul 23 '17 at 21:06
  • $\begingroup$ Thank you HyperBean and Johnathan Gross, it makes more sense now. $\endgroup$ – Rocket Hack Jul 23 '17 at 21:14
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    $\begingroup$ @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. $\endgroup$ – dmckee Jul 23 '17 at 21:23
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It is easier to see when you use k in eV per K which is $$ 8.6173303(50)×10^{-5} \ \mathrm{\frac{eV}{K}} $$

Wikipedia is simply using E = kT

You should always care about your units. Numbers have no physical meaning without them.

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