Question

Based on Free electron theory, the Hall coefficient for Sodium having BCC structure is : 
(Given, side of cube, \(a = 4.28\AA\)) 
 

• A. $3.86 \times 10^{-10} \text{m}^3\text{C}^{-1}$
• B. $-2.45 \times 10^{-10} \text{m}^3\text{C}^{-1}$
• C. $2.45 \times 10^{-8} \text{m}^3\text{C}^{-1}$
• D. $-1.96 \times 10^{-9} \text{m}^3\text{C}^{-1}$

Hint:

For alkali metals (Group 1 like Na, K), the Hall coefficient is always negative (because electrons dominate) and usually on the order of $10^{-10} \text{ m}^3/\text{C}$. This instantly eliminates positive options and vastly differing magnitudes. Remember BCC = 2 atoms/cell, FCC = 4 atoms/cell.

Answer 1

The Correct Option is B

Step 1: Concept:
The question asks for the theoretical Hall coefficient ($R_H$) of Sodium metal using Drude's free electron model. The Hall coefficient relies entirely on the carrier concentration (number of conduction electrons per unit volume).

Step 2: Key Formula or Approach:
The formula for the Hall coefficient is:
\[ R_H = -\frac{1}{ne} \]
Where:
- $n$ is the electron number density (electrons/m$^3$).
- $e$ is the elementary charge ($1.6 \times 10^{-19} \text{ C}$).
- The negative sign indicates that the majority carriers are electrons.
First, we must calculate $n$. Sodium is in Group 1, so it provides 1 valence electron per atom.
Since it has a BCC (Body-Centered Cubic) lattice, there are 2 atoms per unit cell.
Thus, $n = \frac{Z}{a^3}$, where $Z=2$ and $a$ is the lattice parameter.

Step 3: Step-by-step Explanation:

• Calculate the volume of the unit cell:
$a = 4.28 \text{ \AA} = 4.28 \times 10^{-10} \text{ m}$
$V = a^3 = (4.28 \times 10^{-10})^3 \text{ m}^3$
$V \approx 78.39 \times 10^{-30} \text{ m}^3$

• Calculate the number density ($n$):
$n = \frac{2 \text{ electrons}}{78.39 \times 10^{-30} \text{ m}^3}$
$n \approx 0.02551 \times 10^{30} \text{ m}^{-3} = 2.551 \times 10^{28} \text{ m}^{-3}$

• Calculate the Hall coefficient ($R_H$):
$R_H = -\frac{1}{(2.551 \times 10^{28})(1.6 \times 10^{-19})}$
$R_H = -\frac{1}{4.0816 \times 10^9}$
$R_H \approx -0.245 \times 10^{-9} \text{ m}^3\text{C}^{-1}$
$R_H \approx -2.45 \times 10^{-10} \text{ m}^3\text{C}^{-1}$

Step 4: Final Answer:
The calculated Hall coefficient is $-2.45 \times 10^{-10} \text{ m}^3\text{C}^{-1}$, which matches option (B).