Question

Lead atoms occupy a volume of 3 x 10-29 m3. Each atom contributes two free electrons. Calculate the Fermi velocity of lead.

Answer 1

Answer: The Fermi velocity of lead is 64.4 km/s.

Explanation:

To calculate the Fermi velocity, we use the equation:

$V_f=\frac{h}{2\pi m_e}(\frac{3\pi^2N}{V})^{1/3}$

where,

h = Planck's constant = $6.62\times 10^{-34}Js$

$m_e$ = mass of electron = $9.1\times 10^{-31}kg$

N = Number of atoms present in per volume of atom multiplied by number of electrons present in given atom = $\frac{2\times N_A\times V}{M}$

$N_A$ = Avogadro's number = $6.022\times 10^{26}mol^{-1}$    (When the mass is in kilograms)

V = Volume = $3\times 10^{-29}m^3$

M = molecular weight of lead = 207.2 g/mol

Putting values in above equation, we get:

$V_f=\frac{6.62\times 10^{-34}}{2\times 3.14\times (9.1\times 10^{-31})}(\frac{3\times (3.14)^2\times (2\times 6.022\times 10^{23}\times 3\times 10^{-29})}{3\times 10^{-29}\times 207.2})^{1/3}$

$V_f=0.0644\times 10^6m/s=64.4km/s$     (Conversion factor: 1 km = 1000 m)

Hence, the Fermi velocity of lead is 64.4 km/s