Answer
n = 7.62 x $10^{22}$ $cm^{-3}$ = 7.62 x $10^{28}$ $m^{-3}$
Work Step by Step
Given:
number of free electrons per silver atom = 1.3
electrical conductivity of Ag = 6.8 x $10^{7}$ $(Ωm)^{-1}$
density ($p'_{Ag}$) = 10.5 $\frac{g}{cm^{3}}$
atomic weight of Ag = 107.87 $\frac{g}{mol}$
Required:
number of free electrons per cubic meter of Ag
Solution:
As defined in Equation 4.2,
n= 1.3 [$\frac{p'_{Ag}N_{A}}{A_{Ag}}]$
n = 1.3 [$\frac{(10.5 \frac{g}{cm^{3}})(6.022 \times 10^{23} atoms/mol)}{(107.87 g/mol)}$
n = 7.62 x $10^{22}$ $cm^{-3}$ = 7.62 x $10^{28}$ $m^{-3}$
