Answer to Question #103565 in Molecular Physics | Thermodynamics for Ajay

Question

A box of volume 1 cm3
contains
5.2×10^21
electrons. Calculate their Fermi
momentum and Fermi energy.
Take: me=9.1×10^(–28) g
mn=1.67×10^(–24) g
h=6.62×10^(–27) ergs

Answer 1

As per the question,

The volume of the box=$1cm^3$

Number of electrons $n=5.2\times 10^{21}$

$n=\dfrac{N}{V}=\dfrac{5.2\times 10^{21}}{1\times 10^{-6}}$ =$5.2\times10^{27}$

$E_F=\dfrac{h^2}{8m}(\dfrac{3n}{\pi})^{2/3}=\dfrac{(6.62\times 10^{-34})^2}{8\times 9.1×10^{–31}}(\dfrac{3\times 5.2\times10^{27}}{\pi})^{2/3}$

$E_{F}=\dfrac{43.8244\times2.08\times3.0014\times10^{18-68}}{2.145\times72.8\times10^{-31}}$

$E_{F}=\dfrac{273.6\times10^{-50}}{156.2\times10^{-31}}$

$E_{F}=1.75\times10^{-19}J$

SO, Fermi momentum $P_{F}=\sqrt{2mE_F}$

$P_{F}=\sqrt{2\times9.1\times 10^{-31}\times1.75\times 10^{-19}}=\sqrt{31.85\times 10^{50}}$

$P_{F}=5.6\times 10^{25}kg-m/sec$