Question

an element atomic mass 50 having fcc structure has a density of 5.7 g/cm^3. What is the edge length of the unit cell?​

Answer 1

Answer:

a3×Z×MN0×ρ=4×606.02×1023×6.23

a=400 pm

Explanation:

Answer 2

Answer:-

$\red{\bigstar}$ Edge length $\large\leadsto\boxed{\rm\pink{389 \: pm}}$

• Given:-

Atomic mass of an element 50 with fcc structure and density 5.7 g/cm³.

• To Find:-

Edge length of the unit cell.

• Solution:-

We know,

$\underline{\boxed{\sf\purple{Density \: of \: unit \: cell = \dfrac{z \times M}{N \times a^3}}}}$

here,

$\bf{\rho}$ = Density of unit cell

z = Number of atoms present in one unit cell

M = Atomic mass

N = Avogadro's number

a = Edge length

• Substituting in the values:-

➙ $\sf (a)^3 = \dfrac{z \times M}{N \times \rho}$

➙ $\sf (a)^3 = \dfrac{4 \times 50}{6.02 \times 10^{23} \times 5.7}$

➙ $\sf (a)^3 = \dfrac{200}{6.02 \times 10^{23} \times 5.7}$

➙ $\sf (a)^3 = \dfrac{200}{34.314 \times 10^{23}}$

➙ $\sf (a)^3 = 5.89 \times 10^{-23}$

➙ $\sf (a)^3 = 58.9 \times 10^{-24}$

➙ $\sf a = \sqrt{58.9 \times 10^{-24}}$

➙ $\bf a = 3.89 \times 10^{-8} \: m$

➙ $\bf a = 389 \times 10^{-10} \: m$

We know,

$\boxed{\bf\red{1 \: pico meter = 10^{-10}m}}$

Hence,

★ $\large\boxed{\bf\green{389 \: pm}}$

Therefore, the edge length of the unit cell is 389 pm.