Answer to Question #175038 in Mechanical Engineering for Varun

Question #175038

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.


1
Expert's answer
2021-04-01T01:09:23-0400

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"


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