JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Solid State

The density (in

g cm^{- 3}

) of the metal which forms a cubic close packed (ccp) lattice with an axial distance (edge length) equal to 400 pm is ____.

Use: Atomic mass of metal = 105.6 amu and Avogadro's constant

= 6 \times 1 0^{23} mol^{- 1}

Answer: 11

Solution

For a cubic close packed (ccp or fcc) structure:

Number of atoms per unit cell, Z = 4
Edge length, a = 400 pm = 400 × 10⁻¹⁰ cm = 4 × 10⁻⁸ cm
Molar mass, M = 105.6 g/mol
Avogadro's number, Nₐ = 6 × 10²³ mol⁻¹

Using the density formula:

ρ = \frac{Z \times M}{a ^{3} \times N _{A}}

ρ = \frac{4 \times 105.6}{( 4 \times 1 0 ^{- 8} ) ^{3} \times 6 \times 1 0 ^{23}}

ρ = \frac{422.4}{64 \times 1 0 ^{- 24} \times 6 \times 1 0 ^{23}} = \frac{422.4}{384 \times 1 0 ^{- 1}} = \frac{422.4}{38.4} = 11.00 g/cm^{3}

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Numerical answer

Practise more: all JEE Advanced PYQs on Solid State →

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Solid State

Solution

More JEE Advanced questions on Solid State