An element with density 5.03 g cm^-3 forms a fcc lattice with edge length of $4.3×10^{-8} cm$. The atomic mass of the element is:

60.2 u

The correct answer is Option (1) → 60.2 u

To determine the atomic mass of the element, we can use the formula for the density of a face-centered cubic (fcc) crystal lattice. The relationship between density, atomic mass, and lattice parameters is given by:

\(\rho = \frac{Z \cdot M}{N_A \cdot a^3}\)

Where:

\(\rho\) is the density of the element (g/cm³),

\(Z\) is the number of atoms per unit cell (for an fcc lattice, \(Z = 4\)),

\(M\) is the molar mass (atomic mass) of the element (g/mol),

\(N_A\) is Avogadro's number (\(6.022 \times 10^{23} \, \text{atoms/mol}\)),

\(a\) is the edge length of the unit cell (cm).

Given:\

\(\rho = 5.03 \, \text{g/cm}^3\),

\(a = 4.3 \times 10^{-8} \, \text{cm}\),

\(Z = 4\),

\(N_A = 6.022 \times 10^{23} \, \text{atoms/mol}\).

We need to solve for the atomic mass \(M\). Rearranging the formula for \(M\):

\(M = \frac{\rho \cdot N_A \cdot a^3}{Z}\)

Now, plugging in the values:

\(M = \frac{5.03 \times 6.022 \times 10^{23} \times (4.3 \times 10^{-8})^}{4}\)

\(⇒ M \frac{30.29066 \times 10^{23} \times 79.507 \times 10^{-24}}{4}\)

\(⇒ M = \frac{2408.31 \times 10^{-1}}{4}\)

\(⇒ M =\frac{240.8}{4}\)

\(⇒ M \approx 60.2\)