Question

Calculate the value of the Fermi energy for a specific metal when it contain 2.54*10^28
free electrons per cubic meter (m. = 9.11x10-31 kg and h=6.63x10
(A) 1.15 eV
(B) 3.1 eV
(C) 13.6 eV
(D) 0.50 eV​

Answer 1

Given:

Number of conduction per unit volume is, $n = 2.54\times 10^{28} /m^{3}$

Planck's constant, $h = 6.626\times 10^{-34} J.s$

Mass of electron, $m_{e} = 9.11\times 10^{-31} kg$

Find: We have to calculate the value of the Fermi Energy

Step by Step Solution:

We have the expression of the Fermi Energy,

$E_{F} = \dfrac{h^{2} }{8m_{e} }( \dfrac{3n}{2\pi})^{\dfrac{2}{3} }$

$E_{F} = \dfrac{(6.626\times 10^{-34})^{2} }{8\times9.11\times10^{-31} }(\dfrac{3\times 2.54\times 10^{28} }{2\times 3.14})^{\dfrac{2}{3} }$

$E_{F} = 0.6024\times 10^{-37} ( 1.2133\times10^{28} )^{\dfrac{2}{3}} \\\\E_{F} = 0.6024\times 10^{-37} \times 5.2798\times 10^{18}\\\\ E_{F} = 3.1805\times 10^{-19} eV$

so, the option (B) is correct

Answer 2

Answer:c

Explanation: