# The density of aluminum is 2.70 g/cm3 and its molecular weight is 26.98 g/mole. Calculate the Fermi energy If the experimental value of EF is 12 eV, What is the electron effective mass in aluminum? [Aluminum is trivalent]

solution

Given,

Density of aluminium (ρ) =2.70 gm mole-3

molecular weight of alumunium (m)= 26.98 gm mole-1

a) Fermi energy (Ef)= ?

b) Effective mass = ?

we know that

N = $$\frac{vρN_{A}}{m}$$

where

v = valency of atom

NA = Avogardro’s number

N = $$\frac{3 \times 2.70 \times (6.023 \times 10^{23})}{26.98}$$

= 1.807 x 1023 electron m-3

Now we have relation

$$E_f = \frac{h^2}{2m_e} (3N\pi^2)^{\frac{2}{3}}$$

$$= \frac{\frac{\left ( 6.626 \times 10^{23} \right )^2}{2 \pi}}{2 \times 9.1 \times 10^{-31}} (3 \times 1.807 \times 10^{29} \times \pi^2)^{\frac{2}{3}}$$

=1.867 x 10-18 j

= $$\frac{1.867 x 10^{-18}}{1.6 x 10^{-19}}$$

=11.66 ev

b) Again for me*

$$E_f^{‘} = \frac{h^2}{2m_e^*} (3N\pi^2)^{\frac{2}{3}}$$

$$m_e^* = \frac{h^2}{2E_f^{‘}} (3N\pi^2)^{\frac{2}{3}}$$

$$= \frac{\left ( \frac{6.62 \times 10^{-34}}{2 \pi} \right )^2}{2 \times 12 \times 1.6 \times 10^{-19}} (3 \times 1.807 \times 10^{29} \times \pi^2)^{\frac{2}{3}}$$

= 8.847 x 10-31 kg

∴ me* ≈ 0.97 me

Hence, required fermi energy and electron effective mass in aulmunium are 11.66 eV and 0.97 me respectively.