Question

The density of lead, which has the FCC structure, is 11.36 . The atomic weight of lead is 207.19 . Use Avogadro's number: 6.02210. Calculatethe lattice parameter(Enter your answer to three significant figures.) = 2.75*10^21 the atomic radius of lead(Enter your answer to three significant figures.) =

Answer 1

a

`a_o  = 4.95 *10^(-8) \  cm`

b

`r =  1.7500 *  10^(-8) \  cm`

Step-by-step explanation:

From the question we are told that

The density of lead is
`\rho_l =  11.36 \  g/cm^3`

The atomic weight is
`M_w  =  207.19 \  g/mol`

The Avogadro's number is
`N_a  =  6.022 *  10^(23) \  (atom)/(mol)`

Generally for FCC the number of atom is n = 4

Generally the volume of a unit FCC cell is mathematically represented

`V =  (n  *  M_w )/(N_a  *  \rho_l )`

`V =  (4  *  207.19 )/( 6.022 *  10^(23)  *  11.36 )`

`V = 1.211 5 *10^(-22) \  cm^3`

Generally the lattice parameter is mathematically represented as

`a_o  = \sqrt[3]{V }`

`a_o  = \sqrt[3]{1.211 5 *10^(-22) }`

=>
`a_o  = 4.95 *10^(-8) \  cm`

Generally the radius of the lead is mathematically represented as

`r =  (a_o  *  √(2)  )/(4)`

=>
`r =  (4.95 *10^(-8)  *  √(2)  )/(4)`

=>
`r =  1.7500 *  10^(-8) \  cm`