Question

Calculate the number density of free carriers in silver, assuming that each atom contributes one carrier. The density of silver is \( 10.5 \times 10^3 \, \text{kg/m}^3 \) and the atomic weight is 107.8.

• A. \( 0.585 \times 10^{28} \, \text{carriers/m}^3 \)
• B. \( 58.5 \times 10^{26} \, \text{carriers/m}^3 \)
• C. \( 585.0 \times 10^{27} \, \text{carriers/m}^3 \)
• D. \( 5.85 \times 10^{28} \, \text{carriers/m}^3 \)

Hint:

To calculate the number density of free carriers, divide the material's density by the atomic weight and multiply by Avogadro's number.

Answer 1

The Correct Option is A

To find the number density of free carriers in silver, we begin with the given density and atomic weight:

• Density of silver, \( \rho = 10.5 \times 10^3 \, \text{kg/m}^3 \)
• Atomic weight of silver = 107.8

We can find the number of atoms (hence carriers, as each atom contributes one free carrier) per cubic meter using the formula for number density \( n \):
\[ n = \frac{\text{Number of moles in } 1 \, \text{m}^3 \times \text{Avogadro's number}}{1 \, \text{m}^3} \]The number of moles per cubic meter of silver is given by:
\[ \frac{\rho}{\text{Atomic weight}} = \frac{10.5 \times 10^3}{107.8} = 97.39 \, \text{mol/m}^3 \]Using Avogadro's number, \( N_A = 6.022 \times 10^{23} \, \text{atoms/mol} \), calculate the number density:
\[ n = 97.39 \times 6.022 \times 10^{23} = 5.865 \times 10^{28} \, \text{atoms/m}^3 \]Thus, the number density of free carriers in silver is approximately:
\(\boxed{5.865 \times 10^{28} \, \text{carriers/m}^3}\)
This matches the option \( 0.585 \times 10^{28} \, \text{carriers/m}^3 \) when rounded to three significant figures.

Answer 2

The number density of free carriers in a material can be calculated using the formula: \[ n = \frac{\rho \cdot N_A}{A \cdot M} \] Where:
- \( \rho \) is the density of the material,
- \( N_A \) is Avogadro's number (\( 6.022 \times 10^{23} \, \text{mol}^{-1} \)),
- \( A \) is the atomic weight,
- \( M \) is the molar mass. Substituting the given values: \[ n = \frac{10.5 \times 10^3 \times 6.022 \times 10^{23}}{107.8} \approx 0.585 \times 10^{28} \, \text{carriers/m}^3 \]