Q.(a) An element crystallises in bcc lattice with a cell edge of $3 \times 10^{-8}$ cm. The density of the element is $6.89\ g\ cm^{-3}$. Calculate the molar mass of the element. ($N_A = 6.022 \times 10^{23}\ mol^{-1}$)

(a) Molar mass from bcc lattice data

Key idea: For a bcc unit cell, $Z = 2$ atoms per cell. Density relates molar mass $M$ as:

Step 1: Write the formula and substitute $Z = 2$, $a = 3 \times 10^{-8}\ \text{cm}$, $\rho = 6.89\ \text{g cm}^{-3}$, $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$.

Step 2: Compute $a^3 = 27 \times 10^{-24} = 2.7 \times 10^{-23}\ \text{cm}^3$.

Step 3: Solve for $M$:

$M = \frac{6.89 \times 6.022 \times 10^{23} \times 2.7 \times 10^{-23}}{2}$ …

The key to solving part (a) is connecting the macroscopic property of density to the microscopic arrangement of atoms in the crystal lattice. Density is mass per volume. For a crystal, the mass of one unit cell is the mass of the atoms inside it, and the volume is the cube of the edge length. The number of atoms per unit cell ($Z$) depends on the lattice type — for body-centered cubic (bcc), it's 2 atoms per cell.

Let's work through it step by step.

1. Identify the known quantities.

   Edge length, $a = 3 \times 10^{-8}\ \text{cm}$

   Density, $d = 6.89\ \text{g cm}^{-3}$

   Avogadro's number, $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$

   For bcc lattice, number of atoms per unit cell, $Z = 2$

2. Write the density formula for a crystal.

   The density $d$ is given by:

where $M$ is the molar mass in $\text{g mol}^{-1}$. This formula works because $Z \cdot M$ gives the mass of atoms in one unit cell (since $M$ is mass per mole, and $N_A$ atoms make a mole), and $a^3$ is the volume of the cell.

3. Rearrange to solve for molar mass $M$. Multiply both sides by $a^3 \cdot N_A$ and divide by $Z$:

4. Substitute the values. First, compute $a^3$:

Now plug everything in:

5. Simplify step by step. Notice $10^{-23} \times 10^{23} = 1$, so the powers of ten cancel:

Compute the numerator: $6.89 \times 2.7 = 18.603$, then $18.603 \times 6.022 \approx 112.0$ (let's do it carefully: $18.603 \times 6 = 111.618$, $18.603 \times 0.022 = 0.409266$, sum = $112.027266$). So numerator ≈ $112.03$.

Divide by 2:

As a check, a bcc metal of molar mass ≈ 56 g mol⁻¹ at this density is consistent with iron (Fe, bcc, $a \approx 2.87 \times 10^{-8}$ cm, density $7.87$ g cm⁻³, molar mass $55.85$ g mol⁻¹), so the result is reasonable.

So the molar mass is approximately $56.0\ \text{g mol}^{-1}$.

Now for part (b), we need to understand doping in semiconductors.

6. Recall the principle of doping. …