Question

Calculate the mean free path of electrons in a metal, such as silver, at room temperature form heat capacity and heat conduction measurements. Take EF ¼ 5 eV, K ¼ 4:29 102 J/s m K, and Cel v ¼ 1% of the lattice heat capacity. (Hint: Remember that the heat capacity in (21.8) is given per unit volume!)

Answer 1

Answer:

= 4 * 10-8    = 400 Angstrom

Explanation:

EF = 5 eV, K = 4.29 x 102 J/(s m K), and Cvel = 1% of the lattice heat capacity

K= 1/3 (Cv)*v*l

v is fermi velocity which is equal to $v = (2E_f/m)^{0.5}$

after putting mass of electron as $9.1 * 10^{-31}kg$ and $E_f = 5 eV$ we get $v= 1.33 * 10^6 m/s$

$C_v$ is 1% of lattice heat capacity

Heat Capacity of Aluminium is $0.897 J g^{-1}K^-1$

Density = $2.6989 g \ cm^{-3}$

For  lattice heat capacity you need to use the heat capacity for alimunium given  and then multiply with density to get per unit volume term

Heat Capacity per unit volume =   $0.897 J g^{-1}K^-1$ * $2.6989 g \ cm^{-3}$

$= 2.42 J K^{-1} cm^{-3} \\\\= 2.42* 10^6 J K^{-1} m^{-3}$

Cv = 1% of heat capacity per unit volume

$=0.01 * 2.42* 10^8 J K^{-1} m^{-3} \\\\= 2.42* 10^4 J K^{-1} m^{-3}$

Putting values in this equation K= 1/3 (Cv)*v*l

$l = 3K/(C_v * v )\\\\ = 3 * 4.29 * 10^2 / (2.42* 10^4 * 1.33 * 10^6 )$

$= 4 * 10^{-8 }$

  = 400 Angstrom