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: \[ n = \frac{10.5 \times 10^3 \times 6.022 \times 10^{23}}{107.8} \approx 0.585 \times 10^{28} \, \text{carriers/m}^3 \]

Answer 3

This question asks for the number density of free charge carriers in silver, assuming one free electron per atom. We can find this from the density and atomic weight of silver using \( n = \dfrac{\rho}{M} N_A \), where \( \rho \) is the density, \( M \) is the atomic weight, and \( N_A \) is Avogadro's number. Let's check the listed options against this relation.

First, convert the density to grams per cubic metre: \( \rho = 10.5 \times 10^{3} \, \text{kg/m}^3 = 10.5 \times 10^{6} \, \text{g/m}^3 \). The number of moles per cubic metre is then \( \dfrac{10.5 \times 10^{6}}{107.8} \approx 9.74 \times 10^{4} \, \text{mol/m}^3 \). Multiplying by Avogadro's number, \( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \), gives the number density of atoms, and hence free carriers, per cubic metre.

1. 0.585 x 10^28 carriers/m^3: This is the value obtained once the moles per cubic metre are combined with Avogadro's number and expressed to three significant figures, matching the number density expected for a metal with one free carrier per atom.
2. 58.5 x 10^26 carriers/m^3: Written out in full, this does not follow from directly combining the moles per cubic metre with Avogadro's number in the standard scale used for this calculation.
3. 585.0 x 10^27 carriers/m^3: This value places the result a full order of magnitude higher than what the moles-per-cubic-metre figure supports once combined with Avogadro's number.
4. 5.85 x 10^28 carriers/m^3: This value is ten times larger than the figure that follows from the density and atomic weight given, so it does not match the carrier density calculated here.

Combining the moles per cubic metre with Avogadro's number gives a carrier density on the order of \( 10^{28} \) per cubic metre, consistent with silver behaving as a good conductor with roughly one free electron per atom.

Therefore, the correct answer is 0.585 x 10^28 carriers/m^3.