Answer:
The correct answer is 305 pm
Explanation:
The density of a unit cell (d) is given by the following equation:
[tex]d= \frac{zM}{a^{3}N_{a} }[/tex]
Where M is the molar mass, Nₐ is the Avogadro number (6.023 x 10²³ atom/mol), a is the edge of unit cell and z is the number of atoms per unit cell. We have to determine the edge of the unit cell (a) and we know the molar mass from the Periodic Table (M= 50.9415) and the density (d= 5.96 g/cm³). For BCC, z= 2 atom. So, we introduce the data in the equation and we determine a:
⇒ [tex]a= \sqrt[3]{\frac{zM}{dN_{a} } }[/tex]
[tex]a= \sqrt[3]{\frac{(2atom) (50.9415g/mol)}{(5.96g/cm3)(6.023x10^{23}) } }[/tex]
a= 3.05 x 10⁻⁸ cm= 3.05 x 10⁻¹⁰ m
We need the value in pm, so we convert from m to pm (1 pm= 10⁻¹²m):
3.05 x 10⁻¹⁰m x 1 pm/10⁻¹²m= 305.03 pm= 305 pm
