Question

We can look at the band structure of an element to get an idea of how many electronsper atom participate in conduction. We can also determine this numberbased on the free electron density, the atomic mass, and the mass density of thematerial.Calculate the free electron density of lithium. The Fermi energy is 4.72 eV.

Answer 1

Answer:

the free electron density of lithium is 2.3 × 10²⁸ /m³

Explanation:

The Fermi energy is 4.72 eV.

The Fermi energy (highest filled orbital energy ) of free electron is

$\varepsilon_f=\frac{h^2}{8m_e} (\frac{3N}{\pi V} )^{2/3}$

Rearrange the above equation for free electron density of lithium

$\frac{N}{V} = \frac{\pi }{3} (\frac{8m_e \varepsilon_f}{h^2} )^{3/2}$

where,

h = 6.626 × 10⁻³⁴J.s

$m_e$ = 9.11 × 10⁻³¹kg

$\varepsilon_f$ = 4.72eV

$= \frac{\pi }{3} (\frac{8(9.11\times10^{-31})(4.72\times1.6\times10^{-19})}{6.626\times10^{-34}} )\\\\= 2.3\times10^{28} /m^3$

Thus, the free electron density of lithium is 2.3 × 10²⁸ /m³