Question

The density of lead is 11.35 g cm^-3 and the metal crystallizes with fcc unit cell estimate the radius of lead atom (atomic mass of lead = 207g mol^-1 , avogadros no. Is 6.02 × 10^23 mol^-1

Answer 1

Answer:

The radius of the lead atom that crystallizes with fcc unit cell is  $1.75$  × $10^{-8}$ cm·

Explanation:

Given that,

The density of the lead metal  $=$  $11.35\ g\ cm^-^3$

Mass of the lead metal  $=$  $207\ g\ mol^-^1$

Avogadro Number  $=$  $6.02$ × $10^{23} mol^-^1$

Also given that,

The metal lead crystallizes with fcc unit cell·

So, to find out the radius of the lead atom,

First, we need to calculate the volume of the fcc unit cell·

The equation that relates the volume of the fcc unit cell with the density of the lead metal is,

    $d\ =\ \frac{Z\ M}{a^3\ \ N_A\ }$

where,

d   $=$ density of the metal

Z    $=$  No of atoms present in one unit cell

In fcc, z  $=$ $4$

M   $=$  mass of the single metal atom

$a^{3}$   $=$  The volume of the unit cell

$N_A$  $=$  Avogadro Number

Therefore,

  ⇒     $d\ =\ \frac{Z\ M}{a^3\ \ N_A\ }$

  ⇒     $a^{3}\ =\ \frac{Z\ M}{d\ N_A}$

  ⇒     $a^{3}\ =\ \frac{4\ *\ 207\ g\ mol^-^1}{11.35\ g\ cm^-^3\ *\ 6.02\ *\ 10^{23}\ mol^-1 }$

  ⇒     $a^3\ =\ 1.211$ × $10^{-22}\ cm^3$

  ⇒      $a\ =\ 4.95$ ×$10^{-8}\ cm$

Now,

The relation that connects the radius of a lead atom $(\ r\ )$ with the 'a' for the fcc is,

      $a$√$2$ $=\ 4\ r$

⇒    $r\ =\ \frac{a\ \sqrt{2} }{4}$

⇒    $r\ =\ \frac{4.95\ *\ 10^{-8}\ *\ \sqrt{2} }{4}$

⇒    $r\ =\ 1.75$ × $10^{-8}$ cm

Therefore, the radius of the lead atom that crystallizes with fcc unit cell is  $1.75$  × $10^{-8}$ cm·