Question

An element ‘X’ having atomic mass 60 has density 6.23 g cm⁻³. The edge length of its unit cubic cell is 400 pm. (N_A = 6.02 × 10²³ mol⁻¹).

What is the radius of an atom of this element?

विकल्प

• 210.5 pm

• 344.4 pm

• 141.4 pm

• 115.3 pm

Answer 1

141.4 pm

Explanation:

Given: Atomic mass (M) = 60 g/mol

Density (ρ) = 6.23 g/cm³

Edge length (a) = 400 pm = 4.00 × 10⁻⁸ cm

Avogadro’s number (N_A) = 6.02 × 10²³ mol⁻¹

We need to find the radius of the atom, so we first determine the type of unit cell (Z) using the density formula.

`rho  = (Z xx M)/(a^3 xx N_A)`

`Z = (rho xx a^3 xx N_A)/M`

`Z = (6.23 xx (4 xx 10^-8)^3 xx 6.02 xx 10^23)/60`

`Z = (6.23 xx 64 xx 10^-24 xx 6.02 xx 10^23)/60`

`Z = (6.23 xx 385.28 xx 10^-1)/60`

`Z = (2400.2 xx 10^-1)/60`

`Z = 240.02/60`

Z = 4

∴ This is a face-centred cubic (fcc) unit cell.

For fcc,

`a = (4 r)/sqrt 2`

`r = (a xx sqrt 2)/4`

`r = (400 xx 1.414)/4`

`r = 565.6/4`

r = 141.4 pm