Consider the bcc unit cells of the solids 1 and 2 with the position of atoms as shown below. The radius of atom B is twice that of atom A. The unit cell edge length is 50% more in solid 2 than in 1. What is the approximate packing efficiency in solid 2?

Step 1: Determine the Relationship Along the Body Diagonal

In a body-centered cubic (BCC) unit cell, the constituent atoms touch along the body diagonal of the cube. Let the edge length of Solid 2 be $$a_2$$. The length of its body diagonal is equal to $$\sqrt{3}a_2$$.

According to the diagram for Solid 2:

• Corner sites are occupied by atoms of type A, contributing their radius $$r_A$$.
• The body-center site is occupied by a single larger atom of type B, contributing its full diameter $$2r_B$$.

Therefore, along the body diagonal:

$$\sqrt{3}a_2 = 2r_A + 2r_B$$

We are given that the radius of atom B is twice that of atom A ($$r_B = 2r_A$$). Substituting this in:

$$\sqrt{3}a_2 = 2r_A + 2(2r_A) = 6r_A$$

Solving for the unit cell edge length $$a_2$$ in terms of $$r_A$$:

$$a_2 = \frac{6r_A}{\sqrt{3}} = 2\sqrt{3}r_A$$

Step 2: Calculate the Total Volume of Atoms in Solid 2

Let's calculate the effective number of atoms and their volumes contained inside the unit cell:

• Atoms of type A: 8 corners $$\times \frac{1}{8} = 1$$ effective atom of A.
• Atoms of type B: 1 center $$\times 1 = 1$$ effective atom of B.

The total volume occupied by the spheres ($$V_{\text{atoms}}$$) is:

$$V_{\text{atoms}} = \left(1 \times \frac{4}{3}\pi r_A^3\right) + \left(1 \times \frac{4}{3}\pi r_B^3\right)$$

Substitute $$r_B = 2r_A$$:

$$V_{\text{atoms}} = \frac{4}{3}\pi r_A^3 + \frac{4}{3}\pi(2r_A)^3 = \frac{4}{3}\pi r_A^3 (1 + 8) = 12\pi r_A^3$$

Step 3: Calculate the Total Volume of the Unit Cell

The total volume of the cube ($$V_{\text{cell}}$$) is given by $$(a_2)^3$$:

$$V_{\text{cell}} = (2\sqrt{3}r_A)^3 = 24\sqrt{3}r_A^3$$

Step 4: Compute the Packing Efficiency

Packing efficiency is the percentage of total space occupied by the spheres:

$$\text{Packing Efficiency} = \frac{V_{\text{atoms}}}{V_{\text{cell}}} \times 100\%$$

$$\text{Packing Efficiency} = \frac{12\pi r_A^3}{24\sqrt{3}r_A^3} \times 100\% = \frac{\pi}{2\sqrt{3}} \times 100\%$$

Substituting the values $$\pi \approx 3.1416$$ and $$\sqrt{3} \approx 1.732$$:

$$\text{Packing Efficiency} \approx \frac{3.1416}{3.464} \times 100\% \approx 90.7\%$$

Conclusion:

The calculated value yields an approximate space occupancy of roughly $$90\%$$, which matches Option A perfectly.

Answer: Option A — 90 %