Question

Calculate the number of free electrons per cubic meter for some hypothetical metal, assuming that there are 1.3 free electrons per metal atom. The electrical conductivity and density are 6.8 × 107 (Ω-m)-1 and 10.5 g/cm3, respectively, and its atomic weight is 107.87 g/mol. Use scientific notation.

Answer 1

Answer:

The number of free electrons per cubic meter is $7.61\times 10^{28}\ m^{-3}$

Explanation:

It is given that,

The number of free electrons per cubic meter is, 1.3

Electrical conductivity of metal, $\sigma=6.8\times 10^7\ \Omega^{-1}m^{-1}$

Density of metal, $\rho=10.5\ g/cm^3$

Atomic weight, A = 107.87 g/mol

Let n is the number of  free electrons per cubic meter such that,

$n=1.3\ N$

$n=1.3(\dfrac{\rho N_A}{A})$

Where

$\rho$ is the density of silver atom

$N_A$ is the Avogadro number

A is the atomic weight of silver

$n=1.3\times (\dfrac{10.5\ g/cm^3\times 6.02\times 10^{23}\ atoms/mol}{107.87\ g/mol})$

$n=7.61\times 10^{22}\ cm^{-3}$

or

$n=7.61\times 10^{28}\ m^{-3}$

Hence, this is the required solution.