Question

If electrons are treated as distinguishable particles, at what temperature would they have an average energy of 5.5 eV? ( use kB=1.3806×10−23 J/K)

Answer 1

According to the famous equation, we have got:

$$E=\frac32k_BT,\\\space\\ T=\frac{2E}{3k_B}=\frac{2·(5.5·1.602·10^{-19}\text{ J/eV})}{3·1.3806·10^{−23}}= 42551\text{ K}.$$