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Solve the following.
An element with density 10 g cm⁻³ forms a cubic unit cell with edge length of 3 × 10⁻⁸ cm. What is…
An element with density $10\text{ g cm}^{-3}$ forms a cubic unit cell with edge length of $3 \times 10^{-8}\text{ cm}$. What is the nature of the cubic unit cell if the atomic mass of the element is $81\text{ g mol}^{-1}$?
Answer
Answer
Formula used: Density of a unit cell: rho = z*M/(a**3*NA) with M = 81 g mol^-1, a = 3e-8 cm, NA = 6.022e23 mol^-1, rho = 10 g cm^-3, so z = 2.007
$\rho = \dfrac{z M}{a^3 N_A}$, so $z = \dfrac{\rho a^3 N_A}{M}$
$z = \dfrac{10 \times (3 \times 10^{-8})^3 \times 6\cdot022 \times 10^{23}}{81} = 2\cdot01 \approx 2$
Final answer: 2 atoms per unit cell