Silver metal crystallises with a face centred cubic lattice. The length of the unit cell is found to be `3.0xx10^(-8)`cm. Calculate atomic radius and density of silver.
Molar mass of Ag`=108 g mol^(-1), N_(A)=6.02xx10^(23)mol^(-1)`).

Explanation

To solve the problem, we need to calculate the atomic radius and density of silver (Ag) given that it crystallizes in a face-centered cubic (FCC) lattice with a unit cell length of \(3.0 \times 10^{-8}\) cm. The molar mass of silver is \(108 \, \text{g/mol}\) and Avogadro's number is \(6.02 \times 10^{23} \, \text{mol}^{-1}\).

### Step 1: Calculate the Atomic Radius

In a face-centered cubic (FCC) lattice, the relationship between the atomic radius (R) and the unit cell length (a) is given by the formula:

\[
R = \frac{a}{2\sqrt{2}}
...