Question

Calculate the probability for an electron to be found at an energy of (EF + 2kBT)
in a metal

Answer 1

Electrons are particles with half-integer spin. And they obey Fermi-Dirac statistics. The function of Fermi-Dirac distribution is written as follows: 

$P(E,T)=\dfrac{1}{exp( \dfrac{E-E_F}{kT})}$

where $P(E,T)$ - the probability that the electron occupies an energy level $E$ , above or below the Fermi level $E_F$   

Then 

$P(E_F+2k_BT,T)= \dfrac{1}{1+exp(\dfrac{E_F+2k_BT-E_F}{k_BT})}=\dfrac{1}{e^2}=0.119 \newline in \ percentage \ P(E_F+2k_BT,T) =11.9\%$