Ncert Exercises & Solutions

2.35 If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.

1 m = 100 cm
1 cm = $10^{- 2}$ m
Length of the scale = 20 cm
= 20 × $10^{- 2}$ m
Diameter of a carbon atom = 0.15 nm
= 0.15 × $10^{- 9}$ m
One carbon atom occupies 0.15 × $10^{- 9}$ m.
Number of carbon atoms that can be placed in a straight line

$= \frac{20 x 10^{- 2} m}{0.15 x 10^{- 9} m}$
$= 133.33 x 10^{7}$
$= 1.33 x 10^{9}$

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