Question

The temperature dependence of Fermi energy is expressed as:

EF(T)=EF₀
​
[1− π²/12(kT/EF₀)]

where the Fermi energy at T=0 K is given by:
EF₀=(h²/8mₑ)(3n/n)²/³
Consider, the density of gold (Au) is 19.3 gm cm⁻³ and the molar atomic mass is 197 gm/mol. Every gold atom has one valence electron. Assume, melting temperature of gold (Au) is 1337 K and at a specific temperature(T), Ef(T) = 0.99 EFO
Calculate the Fermi energy and Fermi velocity of gold (Au) at T = 0 K.

Answer 1

The Fermi energy of gold at 0 K is approximately 7.09 eV and the Fermi velocity is about 1.39×10¶ m/s, calculated from basic principles using the electron density derived from the density and molar mass of gold.

Step-by-step explanation:

To calculate the Fermi energy and Fermi velocity of gold (Au) at T = 0 K, we start by obtaining the electron density (n) from the given density and molar mass. One mole of Au has Avogadro's number of atoms, each contributing one electron, hence:

n = (density/molar mass) × Avogadro's number

= (19.3 g/cm³ / 197 g/mol) × 6.022×10³³ mol¹
= 5.90 × 10³³ electrons/m³

Using the expression for EF₀, we get:

EF₀ = (h²/8mₑ)(3n/pi)²/³
= [(6.626×10³´ J.s)² / (8×9.109×310³± kg)] × [(3×5.90×10³³ electrons/m³) / pi]²/³
= 7.09 eV (rounded as per the provided table reference).

The Fermi velocity, vF, can then be calculated using the relation:

vF = sqrt(2EF₀/mₑ)
= sqrt(2×7.09 eV / 9.109×310³± kg)
= 1.39×10¶ m/s

Thus, the Fermi energy of gold at 0 K is approximately 7.09 eV, and its Fermi velocity is roughly 1.39×10¶ m/s.