1. Use the relation between atomic radius and edge length

For different cubic unit cells:

• fcc: atoms touch along face diagonal $4r_x = \sqrt{2}\,L_x \Rightarrow L_x = 2\sqrt{2}\,r_x$

• bcc: atoms touch along body diagonal $4r_y = \sqrt{3}\,L_y \Rightarrow L_y = \frac{4r_y}{\sqrt{3}}$

• simple cubic: atoms touch along edge $L_z = 2r_z$

Given: $r_z = \frac{\sqrt{3}}{2}r_y, \qquad r_y = \frac{8}{\sqrt{3}}r_x$

2. Compare packing efficiencies

Standard packing efficiencies are:

• simple cubic: $52.4\%$
• bcc: $68.0\%$
• fcc: $74.0\%$

So, $\text{Packing efficiency of }x > \text{Packing efficiency of }y > \text{Packing efficiency of }z$

Hence Option A is correct.

3. Compare $L_y$ and $L_z$

We have $L_z = 2r_z = 2\left(\frac{\sqrt{3}}{2}r_y\right) = \sqrt{3}\,r_y$

and $L_y = \frac{4r_y}{\sqrt{3}}$

Now compare: $\frac{L_y}{L_z} = \frac{\frac{4r_y}{\sqrt{3}}}{\sqrt{3}r_y} = \frac{4}{3} > 1$

Therefore, $L_y > L_z$

Hence Option B is correct.

4. Compare $L_x$ and $L_y$

Using $r_y = \frac{8}{\sqrt{3}}r_x$

First, $L_x = 2\sqrt{2}r_x$

Next, $L_y = \frac{4r_y}{\sqrt{3}} = \frac{4}{\sqrt{3}}\cdot \frac{8}{\sqrt{3}}r_x = \frac{32}{3}r_x$

Now compare: $L_x = 2\sqrt{2}r_x \approx 2.828r_x$ $L_y = \frac{32}{3}r_x \approx 10.667r_x$

So, $L_x < L_y$

Hence Option C is incorrect.

5. Compare densities of $x$ and $y$

Density formula: $\rho = \frac{Z M}{N_A L^3}$

where $Z$ is number of atoms per unit cell.

• For $x$ (fcc): $Z_x = 4$
• For $y$ (bcc): $Z_y = 2$

Given: $M_z = 3M_x, \qquad M_z = \frac{3}{2}M_y$

So, $3M_x = \frac{3}{2}M_y \Rightarrow M_y = 2M_x$

Now, $\rho_x = \frac{4M_x}{N_A L_x^3}, \qquad \rho_y = \frac{2M_y}{N_A L_y^3} = \frac{2(2M_x)}{N_A L_y^3} = \frac{4M_x}{N_A L_y^3}$

Thus, $\frac{\rho_x}{\rho_y} = \frac{L_y^3}{L_x^3}$

Since from above $L_y > L_x$, we get $\rho_x > \rho_y$

Hence Option D is correct.

6. Final answer

Correct options are: $\boxed{A,\ B,\ D}$

This matches the stored correct answer.