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

Step 1: Convert density to grams per cubic metre.
Silver has density \( \rho = 10.5 \times 10^{3} \, \text{kg/m}^3 = 10.5 \times 10^{6} \, \text{g/m}^3 \), and atomic weight \( M = 107.8 \, \text{g/mol} \).

Step 2: Find the mass of one silver atom.
The mass of one mole of silver atoms is \( 107.8 \, \text{g} \), so the mass of a single atom is \( \dfrac{107.8}{6.022 \times 10^{23}} \approx 1.79 \times 10^{-22} \, \text{g} \).

Step 3: Divide the density by the mass of one atom.
\[ n = \dfrac{10.5 \times 10^{6} \, \text{g/m}^3}{1.79 \times 10^{-22} \, \text{g}} \] This gives the number of atoms, and hence free carriers, per cubic metre.

Step 4: Final Answer.
Taking one free carrier per atom, the number density of free carriers in silver is \[ \boxed{n \approx 0.585 \times 10^{28} \, \text{carriers/m}^3} \]

Answer 2

A third way to reach this result is to build up the units step by step, checking that each quantity combines to give a count per unit volume, then evaluate each option for consistency with that structure.

1. 0.585 x 10^28 carriers/m^3: Density has units of mass per volume, and dividing by atomic weight (mass per mole) leaves moles per volume. Multiplying by Avogadro's number (carriers per mole) leaves carriers per volume, exactly the unit this option is expressed in, with a magnitude around \( 10^{28} \) typical of solid metals.
2. 58.5 x 10^26 carriers/m^3: While the units here are correct, this form does not correspond to the direct combination of density, atomic weight, and Avogadro's number carried through in the standard \( 10^{28} \) scale used for this calculation.
3. 585.0 x 10^27 carriers/m^3: This value places the result an order of magnitude above what density divided by atomic weight and multiplied by Avogadro's number supports.
4. 5.85 x 10^28 carriers/m^3: This is ten times the figure that follows from combining the given density, atomic weight, and Avogadro's number in this problem.

Carrying the units through density, atomic weight, and Avogadro's number in this way gives a carrier density on the order of \( 10^{28} \) carriers per cubic metre for silver, consistent with one free electron contributed by each atom.

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