The Fermi energy for lithium is 4.72 eV at T = 0 K. Calculate the number of conductionelectrons per unit volume in lithium. Given mass of the electron = 9.11 X 10-31 Kg andPlanck’s constant = 6.63 X 10

To calculate the number of conduction electrons per unit volume in lithium, we can use the formula for the density of states in three dimensions:

D(E) = (1/2π^2) * (2m/h^2)^(3/2) * sqrt(E)

where:

• D(E) is the density of states
• m is the mass of the electron
• h is Planck's constant
• E is the energy

First, we need to convert the Fermi energy from electron volts (eV) to joules (J) because the other constants are in SI units. We know that 1 eV = 1.6 x 10^-19 J, so:

E = 4.72 eV = 4.72 * 1.6 x 10^-19 J = 7.552 x 10^-19 J

Now we can substitute the values into the formula:

D(E) = (1/2π^2) * (29.11 x 10^-31 kg / 6.63 x 10^-34 Js)^1.5 * sqrt(7.552 x 10^-19 J)

After calculating the above expression, we get the density of states at the Fermi energy. This value represents the number of states per unit volume per unit energy. To get the number of conduction electrons per unit volume, we need to integrate the density of states from 0 to the Fermi energy. However, at T = 0 K, all states up to the Fermi energy are filled, so the number of conduction electrons per unit volume is equal to the density of states at the Fermi energy.

Please note that this is a simplified model and assumes that all the valence electrons in lithium become conduction electrons, which may not be the case in reality.