Answer to Question #241195 in Physics for Lavanya

Question

4.If electrons are treated as
distinguishable particles, at what temperature would they have an average energy of 5.5 eV? (use kB

= 1.3806 x 10^-23 J/K)


a) 8.88 X 10^4 K

b) 6.38 X10^4 K

c) 6.88 X10^4 K

d) 8.38 ×10^4 K

Answer 1

The energy of the free electron gas is given by:

$$E_F=kT$$

So

$$T=E_F/k\\ =5.5*1.6*10^{-19}/1.38*10^{-23}=6.38*10^{4}\:\rm K$$