JEE Advanced 2023 — Paper 2 — Question 30

For metal ' $x$ ' Fcc: Edge length , $\mathrm{a}_{1}=\mathrm{L}_{\mathrm{x}}$ For metal ' $y$ ' Bcc: Edge length , $\mathrm{a}_{2}=\mathrm{L}_{\mathrm{y}}$ For metal ' $z$ ' Bcc: Edge length , $\mathrm{a}_{3}=\mathrm{L}_{2}$ $r_{z}=\frac{\sqrt{3}}{2} r_{y}, r_{y}=\frac{8}{\sqrt{3}} r_{x}, M_{z}=\frac{3}{2} M_{y}$ and $M_{z}=3 M_{x}$ For option (A) (i) For FCC $(Z=4)$ metal ' $x$ ', $4 r_{x}=\sqrt{2} L_{x}$

$$\text { P.E }=\frac{\mathrm{Z} \times \frac{4}{3} \pi\left(\mathrm{r}_{\mathrm{x}}\right)^{3}}{\mathrm{a}_{1}^{3}}=\frac{4 \times \frac{4}{3} \pi\left(\mathrm{r}_{\mathrm{x}}\right)^{3}}{\left(\mathrm{~L}_{\mathrm{x}}\right)^{3}}=\frac{4 \times \frac{4}{3} \pi\left(\mathrm{r}_{\mathrm{x}}\right)^{3}}{\left(\frac{4}{\sqrt{2}} \mathrm{r}_{\mathrm{x}}\right)^{3}}=0.24 \pi$$

(ii) For BCC $(Z=2)$ metal ' $y$ ', $4 r_{y}=\sqrt{3} L_{y}$

$$\text { P.E }=\frac{\mathrm{Z} \times \frac{4}{3} \pi\left(\mathrm{r}_{\mathrm{y}}\right)^{3}}{\mathrm{a}_{2}^{3}}=\frac{2 \times \frac{4}{3} \pi\left(\mathrm{r}_{\mathrm{y}}\right)^{3}}{\left(\mathrm{~L}_{\mathrm{y}}\right)^{3}}=\frac{2 \times \frac{4}{3} \pi\left(\mathrm{r}_{\mathrm{y}}\right)^{3}}{\left(\frac{4}{\sqrt{3}} \mathrm{r}_{\mathrm{y}}\right)^{3}}=0.22 \pi$$

(iii) For SC ( $\mathrm{Z}=1$ ) metal ' z ', $2 \mathrm{r}_{\mathrm{z}}=\mathrm{L}_{\mathrm{z}}$

$$\text { P.E }=\frac{\mathrm{Z} \times \frac{4}{3} \pi\left(\mathrm{r}_{\mathrm{z}}\right)^{3}}{\mathrm{a}_{3}^{3}}=\frac{1 \times \frac{4}{3} \pi\left(\mathrm{r}_{z}\right)^{3}}{\left(\mathrm{~L}_{\mathrm{z}}\right)^{3}}=\frac{1 \times \frac{4}{3} \pi\left(\mathrm{r}_{\mathrm{z}}\right)^{3}}{\left(2 \mathrm{r}_{z}\right)^{3}}=\frac{\pi}{6}=0.17 \pi$$

$(\text { P.E })_{\mathrm{FCC}}>(\text { P.E })_{\mathrm{BCC}}>(\text { P.E })_{\mathrm{SC}}$ For option (B)

$$\begin{aligned} & 4 r_{y}=\sqrt{3} L_{y} 2 r_{z}=L_{z} \\& L_{y}=\frac{4 r_{y}}{\sqrt{3}} \\& \frac{L_{y}}{L_{z}}=\frac{4 r_{y}}{\sqrt{3} \times 2 r_{z}}=\frac{2 r_{y}}{\sqrt{3} r_{z}}=\frac{2 r_{y}}{\sqrt{3} \cdot \frac{\sqrt{3}}{2} r_{y}}=\frac{4}{3} \end{aligned}$$

So, $L_{y}>L_{z}$ For option (C)

$$\begin{aligned} & 4 r_{x}=\sqrt{2} L_{x}, 4 r_{y}=\sqrt{3} L_{y} \\& L_{x}=\frac{4 r_{x}}{\sqrt{2}}, L_{y}=\frac{4 r_{y}}{\sqrt{3}} \\& \frac{L_{x}}{L_{y}}=\frac{\sqrt{3} r_{x}}{\sqrt{2} \times 8 / \sqrt{3} r_{x}}=\frac{3}{8 \sqrt{2}} \end{aligned}$$

So, $\mathrm{L}_{\mathrm{x}}<\mathrm{L}_{\mathrm{y}}$ incorrect For option (D) $d_{x}=\frac{4 \times M_{x}}{\left(\frac{4 r_{x}}{\sqrt{2}}\right)^{3} \times N_{A}}$ $d_{y}=\frac{2 \times M_{y}}{\left(\frac{4 r_{y}}{\sqrt{3}}\right)^{3} \times N_{A}}$ $r_{y}=\frac{8}{\sqrt{3}} r_{x}, \frac{M_{x}}{M_{y}}=\frac{1}{2}$ $\frac{\mathrm{d}_{\mathrm{x}}}{\mathrm{d}_{\mathrm{y}}}=\frac{512}{2 \sqrt{2}}=\frac{256}{\sqrt{2}}$ So $\mathrm{d}_{\mathrm{x}}>\mathrm{d}_{\mathrm{y}}$ (correct)