Ncert Exercises & Solutions

2.37 The diameter of zinc atom is 2.6 Å.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.

(a) Radius of zinc atom = Diameter/2

$= \frac{2.6 A}{2}$
$= 1.3 x 10^{- 10} m$
$= 130 x 10^{12} m = 130 pm$

(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

$= \frac{1.6 x 10^{- 2} m}{2.6 x 10^{- 10} m}$
$= 0.6153 x 10^{8} m$
$= 6.153 x 10^{7}$

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