Answer: The Fermi velocity of lead is 64.4 km/s.
Explanation:
To calculate the Fermi velocity, we use the equation:
[tex]V_f=\frac{h}{2\pi m_e}(\frac{3\pi^2N}{V})^{1/3}[/tex]
where,
h = Planck's constant = [tex]6.62\times 10^{-34}Js[/tex]
[tex]m_e[/tex] = mass of electron = [tex]9.1\times 10^{-31}kg[/tex]
N = Number of atoms present in per volume of atom multiplied by number of electrons present in given atom = [tex]\frac{2\times N_A\times V}{M}[/tex]
[tex]N_A[/tex] = Avogadro's number = [tex]6.022\times 10^{26}mol^{-1}[/tex] (When the mass is in kilograms)
V = Volume = [tex]3\times 10^{-29}m^3[/tex]
M = molecular weight of lead = 207.2 g/mol
Putting values in above equation, we get:
[tex]V_f=\frac{6.62\times 10^{-34}}{2\times 3.14\times (9.1\times 10^{-31})}(\frac{3\times (3.14)^2\times (2\times 6.022\times 10^{23}\times 3\times 10^{-29})}{3\times 10^{-29}\times 207.2})^{1/3}[/tex]
[tex]V_f=0.0644\times 10^6m/s=64.4km/s[/tex] (Conversion factor: 1 km = 1000 m)
Hence, the Fermi velocity of lead is 64.4 km/s
