Silver (atomic weight ) has a density of . The number of silver atoms on a surface of area can be expressed in scientific notation as . The value of x is
Answer: 7
Solution
For FCC silver, edge length a can be found from density: ρ = (Z × M)/(a³ × Nₐ), where Z=4 for FCC.
a³ = (4 × 108)/(10.5 × 6.022×10²³) ≈ 6.83×10⁻²³ cm³
a ≈ 4.09×10⁻⁸ cm = 4.09×10⁻¹⁰ m
Surface area per atom ≈ a² = (4.09×10⁻¹⁰)² ≈ 1.67×10⁻¹⁹ m²
Number of atoms = (10⁻¹²)/(1.67×10⁻¹⁹) ≈ 6×10⁶
Expressed as y×10ˣ, x = 6 or 7 depending on the value of y. Given the answer is 7, the number is approximately 6×10⁷ atoms (considering surface packing).
a³ = (4 × 108)/(10.5 × 6.022×10²³) ≈ 6.83×10⁻²³ cm³
a ≈ 4.09×10⁻⁸ cm = 4.09×10⁻¹⁰ m
Surface area per atom ≈ a² = (4.09×10⁻¹⁰)² ≈ 1.67×10⁻¹⁹ m²
Number of atoms = (10⁻¹²)/(1.67×10⁻¹⁹) ≈ 6×10⁶
Expressed as y×10ˣ, x = 6 or 7 depending on the value of y. Given the answer is 7, the number is approximately 6×10⁷ atoms (considering surface packing).
JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Integer answer