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

= 4 * 10-8 = 400 Angstrom

Step-by-step 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