Question

The Fermi energy level for copper is 6.25 eV. Determine the temperature at 

which there is a 1% probability that an energy state 0.30 eV below the Fermi 

energy level will not contain an electron.

Answer 1

The probability that the state is empty, is 

$$1-f(E_V)=1-\dfrac{1}{1+e^{(E_V-E_F)/(k_BT)}}$$

Substitute

$$0.01=1-\dfrac{1}{1+e^{(-0.30\ eV)/(k_BT)}}$$

$$e^{(-0.30\ eV)/(k_BT)}=\dfrac{1}{1-0.01}-1$$

$$e^{(0.30\ eV)/(k_BT)}=99$$

$$\dfrac{0.30\ eV}{k_BT}=\ln99$$

$$T=\dfrac{0.30\ eV}{k_B\ln99}$$

$k_B=8.6173\times10^{-5}\ eV\cdot K^{-1}$

$$T=\dfrac{0.30\ eV}{8.6173\times10^{-5}\ eV\cdot K^{-1}\ln99}$$

$$T=757.62\ K$$