The diameter of a zinc atom is $2.6 \, \mathring{A}$. Calculate $(a)$ the radius of the zinc atom in $pm$ and $(b)$ the number of atoms present in a length of $1.6 \, cm$ if the zinc atoms are arranged side by side lengthwise.

$(a)$ Radius of zinc atom $= \frac{\text{Diameter}}{2} = \frac{2.6 \, \mathring{A}}{2} = 1.3 \, \mathring{A} = 1.3 \times 10^{-10} \, m = 130 \, pm$.
$(b)$ Length of the arrangement $= 1.6 \, cm = 1.6 \times 10^{-2} \, m$.
Diameter of zinc atom $= 2.6 \, \mathring{A} = 2.6 \times 10^{-10} \, m$.
Number of zinc atoms $= \frac{\text{Total length}}{\text{Diameter of one atom}} = \frac{1.6 \times 10^{-2} \, m}{2.6 \times 10^{-10} \, m} \approx 6.153 \times 10^7$ atoms.