Answer:
a
[tex]a_o = 4.95 *10^{-8} \ cm [/tex]
b
[tex]r = 1.7500 * 10^{-8} \ cm[/tex]
Explanation:
From the question we are told that
The density of lead is [tex]\rho_l = 11.36 \ g/cm^3[/tex]
The atomic weight is [tex]M_w = 207.19 \ g/mol[/tex]
The Avogadro's number is [tex]N_a = 6.022 * 10^{23} \ \frac{atom}{mol}[/tex]
Generally for FCC the number of atom is n = 4
Generally the volume of a unit FCC cell is mathematically represented
[tex]V = \frac{n * M_w }{N_a * \rho_l }[/tex]
[tex]V = \frac{4 * 207.19 }{ 6.022 * 10^{23} * 11.36 }[/tex]
[tex]V = 1.211 5 *10^{-22} \ cm^3[/tex]
Generally the lattice parameter is mathematically represented as
[tex]a_o = \sqrt[3]{V } [/tex]
[tex]a_o = \sqrt[3]{1.211 5 *10^{-22} } [/tex]
=> [tex]a_o = 4.95 *10^{-8} \ cm [/tex]
Generally the radius of the lead is mathematically represented as
[tex]r = \frac{a_o * \sqrt{2} }{4}[/tex]
=> [tex]r = \frac{4.95 *10^{-8} * \sqrt{2} }{4}[/tex]
=> [tex]r = 1.7500 * 10^{-8} \ cm[/tex]
