At what temperature do 1.30% of the conduction electrons in lithium (a metal) have energies greater than the Fermi energy EF, which is 4.70 eV? (See Problem 21)

Short Answer

Expert verified

The temperature at which 1.30% of the conduction electrons in lithium have energies greater than the Fermi energy is 472.5 K .

Step by step solution

01

The given data

a) The fraction of conduction electrons having energies greater than the Fermi energy, frac=0.013

b) Fermi energy of lithium, EF=4.70eV

02

Understanding the concept of fraction of conduction electrons

The electrons that have jumped to the conduction band from the valence band and are now free to move within the walls of the substance are known as conduction electrons.

Formula:

The fraction of electrons with energies greater than the Fermi energy is

frac=3kT2EF ……( i )

wherek=8.62×10-5eV/K

03

Calculation of the temperature

Using the given data in equation (i), we get the temperature at which the 1.30% energy is greater than the Fermi energy can be calculated as follows:

T=2fracEF3k=20.0134.70eV8.62×10-5eV/K=472.5K

Hence, the value of the temperature is 472.5 K.

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