Question

The edge length of a unit cell of a metal having molecular weight $75$ g/mol is $5\mathring{A}$ which crystallizes in a cubic lattice. If the density is $2$ g cm${}^{-3}$, then find the radius (in pm) of the metal atom.

[Use $N_A=6\times {10}^{23}$]

• A. $108$
• B. $433$
• C. $250$
• D. $216$

Answer 1

Answer: (D)

$216$

The expression for density is given below:

$\rho =\frac{Z_{\text{eff}}\times M}{a^3\times N_A}$

or $2=\frac{Z_{\text{eff}}\times 75}{(5\times {10}^{-8}{)}^3\times 6\times {10}^{23}}$

On solving, we get $Z_{\text{eff}}\approx 2$.

This value is for bcc structure.

For bcc, $r=\frac{\sqrt{3}}{4}a=\frac{\sqrt{3}}{4}\times 5=2.165\mathring{A}=216.5$ pm

Hence, the correct option is $\text{D}$