Question

Gold (Au) is in the same group on the periodic table as copper (Cu) with one valance electron. Calculate the mobility of the electrons in gold. Assume the density is 19.3 g/cm^3 and the atomic mass is 196.67 g/mol and the conductivity of gold is 4.1 Times 10^7 S/m

Answer 1

given,

density, d = 19.3 g/cm^3

atomic mass, m = 196.67 g/mol

No.of gold atoms per unit volume, n = (density/atomic mass)* Avogadro number

= (19.3/196.67) * 6.03*10^{23}

= 5.9174 *10^{22} electrons/cm^3

= 5.9174*10^{26} electrons/m^3

Relaxation time is

$\tau =\frac{m\rho}{ne^2}=\frac{9.1*10^{-31}*4.1*10^7}{5.9174*10^{28}*(1.6*10^{-19})^2}$

= 2.463*10^{-14} s

The electron mobility is

  $\mu =\frac{e\tau}{m} = \frac{1.6*10^{-19}*2.463*10^{-14}}{9.1*10^{-31}}=4.3305*10^{-3}\:m^2/V-s$