Answer:
The answer to the question is
The approximate atomic radius for silver is 1.445×10⁻⁸ cm
Explanation:
To solve the question we list out the known variables as follows
The density of silver = 10.5 g/cm³
Molar mass of silver = 107.9 g/mol
Number of moles in one cm³ of silver = 10.5/107.9 moles or 0.0973 moles
Avogadro's law states that equal volumes of all substances contain equal number of particles that is one mole of any substance contain 6.022 × 10²³ elementary particles
Therefore one mole of silver contains 6.022 × 10²³ silver atoms
number of moles of silver in a unit cell = 4/6.022 × 10²³ or 6.642 × 10⁻²⁴ mol which has a mass = 6.642 × 10⁻²⁴ mol × molar mass of silver or 7.167× 10⁻²² g
Therefore the volume of a unit cell is given by Volume = mass/density =
7.167× 10⁻²² g/10.5 g/cm³ = 6.83× 10⁻²³ cm³
The diagonal of the face of a unit cell contains four atomic silver radius therefore
That is 4 × silver radius = diagonal of cubic unit cell face
= √2 × ∛(6.83 × 10⁻²³ cm³)
The approximate atomic radius for silver = (1/4) × √2 × ∛(6.83× 10⁻²³ cm³)
= 1.445×10⁻⁸ cm
