If copper, density = 9.0 g/cm^3 and atomic mass 63.5, bears face-centered

if-copper-density-9-0-g-cm-3-and-atomic-mass-63-5-bears-face-centered

If copper, density = 9.0 g/cm^3 and atomic mass 63.5, bears face-centered unit cells then what is the ratio of surface area to volume of each copper atom?

A. 0.0028

B. 0.0235

C. 0.0011

D. 0.0323

I had been asked this question during an online exam.

This interesting question is from Solid State in division Solid State of Chemistry – Class 12

The correct answer is B. 0.0235

Easiest explanation: Density, d of unit cell is given by d = \(\frac{zM}{a^3N_A}\)

Given,

Density, d = 9.0 g/cm^3

Atomic mass, M = 63.5 g/mole

Edge length = a

NA = Avogadro’s number = 6.022 x 10^23

z = 4 atoms/cell

On rearranging the equation for density we get a^3 = \(\frac{zM}{dN_A}\)

Substituting the given values:

a^3 = \(\frac{4 \times 63.5}{9 \times 6.022 \times 10^{23}}\)

Therefore, a = 360.5 pm

 The relation of edge length ‘a’ and radius of particle ‘r’ for FCC packing i.e. a = 2\(\sqrt{2}\)r.

On substituting the value of ‘a’ in the given relation, r = \(\frac{360.5}{2\sqrt{2}}\)=127.46 pm

Now, for spherical particles volume, V = 4πr^3/3 and surface area, S = 4πr^2

Required ratio = S/V=4πr^2/(4πr^3/3) = 3/r (after simplifying)

Thus, S/V = 3/127.46 = 0.0235.