10$$^{-8}$$ cm

$$\rho=2$$ g cm$$^{-3}$$

$$\mathrm{N_A=6\times10^{23}}$$

To Find: Radius of metal atom ($$r$$)

To find the value of $$r$$, we need to find out the type of unit cell. For that, the value of Z is required. On the basis of the value of Z the type of unit cell can be identified and using suitable formula, radius of metal atom will be determined.

Step 1 : Determination of $$Z_{\text {eff }}$$

The formula for calculating $$\mathrm{Z}_{\text {eff }}$$ will be obtained from the formula for density.

The formula for calculating density is,

$$\rho=\frac{\mathrm{Z} \times \mathrm{M}_{w}}{a^{3} \times \mathrm{N}_{\mathrm{A}}}$$

Here, $$\rho$$ is the density of unit cell, Z is the number of atoms present in a unit cell, $$\mathrm{M}_{w}$$ is the molecular weight, $$a$$ is the edge length of a unit cell and $$\mathrm{N}_{\mathrm{A}}$$ is the Avogadro's number.

On rearranging this formula,

$$\mathrm{Z}=\frac{\rho \times a^{3} \times \mathrm{N}_{\mathrm{A}}}{\mathrm{M}_{w}}$$ ...... (i)

Substituting the respective values in equation (i),

$$Z=\frac{2 \times\left(5 \times 10^{-8}\right)^{3} \times\left(6 \times 10^{23}\right)}{75}$$

$$\therefore \quad \mathrm{Z}=1.98 \approx 2$$

The value of Z is 2 for BCC (body centered cubic) unit cell.

Step 1: Determination of radius of metal atom.

For $$\mathrm{BCC}$$, the radius of atom is given by,

$$4 r=\sqrt{3} a$$

Here, $$r$$ denotes the radius of the atom.

$$\begin{array}{ll} \therefore & r=\frac{\sqrt{3}}{4} a \\ \therefore & r=\frac{\sqrt{3}}{4} \times 5=2.165 ~\mathop A\limits^o \\ \therefore & r=216.5 ~\mathrm{pm} \end{array}$$

The radius of metal atom is $$216.5 ~\mathrm{pm}$$.

Solid State

Continue this paper

Similar Solid State PYQs

More practice from this question