Question

The diameter of the zinc atom is $2.6\mathop A\limits^o$. Calculate:
A. The radius of zinc atom in pm and
B. The number of atoms present in a length of 1.6 cm if the zinc is arranged side by side lengthwise.

Answer 1

Radius is defined as the distance from the center to outwards whereas the diameter goes straight through the center. The radius is one-half of the diameter. Mathematically:
$Diameter = 2 \times {\text{Radius}}$

Complete step by step answer:
Here it is given that the diameter of the zinc atom is $2.6\mathop A\limits^o$
For calculating the radius of zinc, we assume the shape of the atom is spherical hence by using the formula given, we calculate the radius of the zinc atom as :
$Diameter = 2 \times {\text{Radius}}$
$2.6 = 2 \times {\text{Radius}}$
$Radius = \dfrac{1}{2} \times 2.6 = 1.3\mathop A\limits^o$
As we know $0.01\mathop A\limits^o = 1pm$
Hence, $1.3\mathop A\limits^o = \dfrac{{1 \times 1.3}}{{0.01}} = 130pm$
To calculate the number of atoms,
Here the length of the arrangement is given 1.6 cm and the diameter of the atom is $2.6\mathop A\limits^o$. As we know 100cm=1 m and $1\mathop A\limits^o$=${10^{ - 10}}m$. Hence we can write the length of arrangement =1.6×10-2 m and the diameter of atom =2.6×10-10m.
So the number of atoms present in the arrangement =$\dfrac{{1.6 \times {{10}^{ - 2}}}}{{2.6 \times {{10}^{ - 10}}}} = 0.6153 \times {10^8}m$

Additional Information: Unit cell is defined as the basic unit of repeating patterns in crystalline solids and there are seven types of unit cell that differ mainly in length of edges and angles between them. The type of Unit cells are:
-Cubic cell
-Tetragonal cell
-Orthorhombic cell
-Monoclinic cell
-Hexagonal cell
-Rhombohedral cell
-Triclinic cell

Note: The atomic radius of an element is the distance from the center of the nucleus to the boundary of the surrounding shell of electron and since there is no well-defined boundary available, therefore we have other radius such as ionic radius, covalent radius or Van der Waals radius.