Question

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

Answer 1

(a) Radius of zinc atom = $\frac{Diameter}{2}$
$=\frac{2.6A}{2}=1.3\times {10}^{-10}m=130\times {10}^{-12}m=130pm$

(b) Length of the arrangement = 1.6 cm
= 1.6 × ${10}^{–2}$ m
Diameter of zinc atom = 2.6 × ${10}^{–10}$ m
∴ Number of zinc atoms present in the arrangement
$\frac{1.6\times {10}^{-2}m}{2.6\times {10}^{-10}m}=0.6153\times {10}^{8}m=6.153\times {10}^7$